SkylineRB26DETT, on 02 August 2010 - 11:40 AM, said:
There are more sectors I keep finding.
Yep

, keep 'em coming, the ones you just posted seem completely unlike "them".
There is a possibility that has just come to my mind.
If the drive was originally partitioned by Vista

or Windows 7, it might have had a "wrong" (from the old "standard" view-point) sector alignment.
I.e. it could have been aligned to the "cluster size" instead of on cylinder border.
Just a guess, but if the recovery partition was made with an older OS, it would have had the "normal" 0/1/1 start and n/254/63 end (in this case 1023/254/63 .i.e. "a suffusion of yellow" since recovery partition is bigger than the CHS limit), and your recovery partition does have this values.
Then comes into play a "standard" Vista

or 7 that aligns partitions differently.
A "normal" first partition starts at 0/1/1 and ends at
n/254/63.
The same if created on an unpatched NT 6/7 will start on different address, aligned with 128 sectors before:
http://www.911cd.net...showtopic=21186
most probably 0/2/3 but normally end on border, like
m/254/63, see here for an example:
http://www.msfn.org/...ic=119963&st=17
Cannot say if non-first partitions would be as well aligned like that, but if they are, then it is possible that the "actual partition" starts not at 20482875 LBA, but rather at 20482875-63+128=20482940.
Then TESTDISK "thought" that partitions were bounded to cylinder borders and when you told it to create the bootsector, it re-created the bootsector in the "wrong" place, also adjusting the partitioning data.
If this is the case, the $MFT should be at sector 6291456-63+128=6291521
This would make sense
IF you did
not (high probability

) go into "Options" and changed "Cylinder Boundary" from the default "Yes" to "No" and the "Allow partial last cylinder" from the default "No" into "Yes".
I do know that the above seems complex and confusing (actually
it is complex and confusing

).
jaclaz
This post has been edited by jaclaz: 03 August 2010 - 03:11 AM