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Originally Posted by Stompy
1. I'm thinking about selling my 30GB iPod, but since I've used it to store random files (backup and otherwise), I want to use DBAN to do a DoD 7-pass wipe on it.
If I do that, will the iPod cease to function because of files required to be there, or will the host software see that it's a newly formatted Fat32 and create everything it needs? In other words: is there anything critical for the iPod's functionality that I need to back up?
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I don’t know what format the drive has to be, I’ve never owned one; however, if you run dban, then reformat the drive with ipod’s software you should be fine.
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Originally Posted by Stompy
2. Non relating to iPods - someone mentioned that even if you wipe a disk with a 7-pass wipe, data is still recoverable. This got a "WTF, no way.. HOW?!" response out of me. Is this possible?
It was explained to me as a sort of "disk write" algorithm and that if someone knew the algorithm of the software used to write those junk bits, it could be reversed. This made no sense because this isn't encryption, or algorithm placed on the data itself.
For example, if you encrypt "apple", it could become "u3*38", and if you know how to undo it, yes, you will get the original value of "apple". But if you have "apple" and make it "00000" .. then "11111" then "8vcjw" - how in the WORLD would you originally know that it contained "apple"?
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Each time you write to a hard drive, it flips the bits on the platter, however, it never truly flips it all the way, some one with the proper tools, and plenty of time can go back several layers of data. if the data is overwritten by all 0's, it is very easy for some one who has the tools to recover the data, that is why you want to do a random write several times, this being said, if you knew exactly what was written with each pass, it would be more likely to recover the data, how ever, it is nearly impossible. Unless you have a government going after you, you don’t need to worry past 7 random passes.
another way of looking at it is to think of each bit is the rounded result of |sin(x)| where x is the angle of the bit, the bit is reported as a 0 if it is close to a 0 and 1 if it is close to a 1, hoe ever, it may be slightly off, this slight discrepancy is detectable, but not by the drive. From this, you can get what the bit used to be, several layers back.