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Start sector Dec to bin

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#1 betrand


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Posted 13 April 2012 - 05:05 PM

I have a USB stick whose Mbr is faulty.
I used a recovery tool. It's a demo so it can't rebuild.
It found the partition I'm looking for.
I says start sector 69. Cool.

I am trying to put that 69 in the binary values of the MBR.
69 in binary is 1000101. The entry for start sector should be 6 digits long.

Where am I wrong?


#2 steve6375


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Posted 13 April 2012 - 11:25 PM

Is there a sector 0 - if so then sector 69 is Logical Block Address 69 which is actually sector number 70 (as sector numbers start at 1)

Max sector number is 63. So you need to have cyl=0 head=1 Sector=x  where x equates to the sector number + 63 (assuming 63 sectors per track geometry).

#3 betrand


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Posted 14 April 2012 - 07:23 PM

Hi, thanks for that.

It seems it's Their image (ActiveRecovery), which starts at 69. So my stick is probably normal (not LBA).
Stick was Fat32, 512/63. 1 partition.
Testdisk doesn't find any mbr (both modes).
I'll try some other tools too.


#4 Wonko the Sane

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Posted 21 April 2012 - 10:11 AM

Where am I wrong?

In several places (not everywhere but here and there) :whistling:.
It is VERY UNlikely that a partition starts on sector 69. (no OS or partitioning tool on earth, on *any* geometry, will ever do that unless "forced to")
There is NO binary to be "input" in a MBR, there are TWO start addresses, one CHS and one LBA, BOTH expressed in HEX that go in the partition table, together with some other data (Partition ID, Active status) and two end addresses.
TESTDISK does not look for any "MBR" at all, it tries to look for bootsectors/PBRs.

Read here:
try the spreadsheets.

Get a disk/hex editor with a template for MBR/partition tables, suggested:


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