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I'm running CentOS 6 on a server that has two 80 GB internal drives, in a software RAID 1. I want to replace this two internal drives with 750 GB internal drives I have available.

How do I do this without rebuilding the system from scratch?

I am thinking of maybe booting up from DVD-ROM and just putting one of the larger drives in while leaving the old drive in there. Then, format the new drive and create partitions that match the smaller drive. Of course, the larger drive with different partition sizes cause there is more space to use) and rsync each partition over. Then repeat this process with putting in the second 750 GB drive and rsync it back from the other 750GB drive.

I'm concerned that I am overlooking something with this method.

share|improve this question
It would be nice if you clarify what SW raid solution is used and if your server is capable to connect two more drives simultaneously. – Serge Oct 21 '12 at 0:22
The server only has slots for two internal drives. An additional two drives via eSATA would be connected if needed. It's a software RAID 1 which is supported by CentOS. Please be specific if you need additional information. Thanks! – Edward Oct 21 '12 at 11:50
Yes, please. I got from the original post that you are running CentOS. I am more interested if you run the LVM on top of this raid. – Serge Oct 21 '12 at 12:44
Not running LVM. – Edward Oct 21 '12 at 12:49
then perhaps the method you described is the optimal one. If you were using LVM you would create the new array from it, then create a PhysicalVolume on that new array, add it to the existing Volume Group and migrate all Physical Extent's off the old array. Then remove the old array PV's from VG. So, nearly zero downtime. Unfortunately this is not an option in your case. – Serge Oct 21 '12 at 13:04

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