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Hello Mac,

Thank you for your time in responding to my post.

chkdsk results:

Drive C: 19,541,504 kb total 1,174,608 kb free 16,384 b in each unit 1,221,344 total 
units 73,413 available 655,360 total mem 593,840 free 

Drive D: 97,659,328kb total disk 65,684,224kb free 32,768 b in each unit 3,051,854 
total units 2,052,632 Available 

Your post stated "You can obtain this cluster size on a FAT32 volume using: A:\>format 
c: /Z:8" is this something I should do now or do I need to wait?

I saw another post of yours stating that a format could be done on a drive in use 
and instructions on how do to so. Just haven't followed up on that yet.

Thanks again, Jeff

Drive C: = 18.63 Gb , 16,384 bytes per allocation unit
Drive D: = 93.14 Gb , 32,768 bytes per allocation unit

These drives are huge and because they are using FAT32 the cluster sizes are very 
high.
I would, personally partition them both, especially as you are less likely to lose 
all of the data on one drive at any one time unless the drive fails completely.

You can only format a drive that is empty. Formatting removes ALL data!

So if your C: drive is OK, and your D: drive is empty you can format your D: drive.

I would start over and reinstall with drives this size. Work out what you need to 
do and partition drive 1. Four drives per physical drive is the maximum.

Primary DOS drive C: set Active. (Mine is only 4·0 Gb)

An Extended Partition which takes up the rest of the drive and three logical drives 
D: (Documents) E: (Email) F: (Music)

Then on drive 2 another Extended DOS partition with four drives, G: H: I: & J:





On drive 1 there is not a problem as it is easy to get four drives under 8Gb so they 
will come out to 4096 bytes anyway.

Drive D: is something to experiment with, which is what I am doing. Four 25Gb drives 
will still have a large cluster size unless you force the smaller size using the 
/Z:8 switch.

Up to you, Jeff.
Mac

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