I have bought a Plextor ssd (m5s-120GB) and I need some help in partitioning it in Ranish Partition Manager for win xp sp3.
Found this thread- viewtopic.php?t=55530
; wherein cmthompson says " ....a program on the Ultimate Boot CD, called Ranish that will create a partition on an arbitrary C/H/S boundary. The correct values to use are 0/16/17, which equals 512 KB....."
Then below CTT replies as "I think MS recommends a 2048-sector alignment (1 MB) and Windows Vista also aligns at 1 MB by default during installation (if you have an unpartitioned drive) so I would also go for this value "just in case" - a 1 MB aligned partition is also 512 KB aligned, but not the other way around; seems that the erase block size for most flash chips is 512KB."
Now my questions:
1. What are the C/H/S values for 1mb alignment (instead of 0/16/17) ?
2. This 0/16/17 is, I suppose, the 1st 3 numbers in RPM i.e. the start of a partition. What should the last 3 numbers in RPM i.e. end of the partition (rather, what should be the last 2 numbers, as the 4th number will depend on size of the partition?)?
3. Has my system (c) partition to be the 1st partition in RPM, compulsorily ( i prefer to have a small dos/fat16 24mb partition as 1st partition in RPM)?
4. After getting the partition offset number (of the partition) in xp, what is the number to divide it with ( so that result is a whole number), so that it confirms I have a 1mb aligned partition?