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From: JW on 16 Apr 2010 05:11
On Thu, 15 Apr 2010 16:50:12 -0700 "Joel Koltner"
<zapwireDASHgroups(a)yahoo.com> wrote in Message id:
<VCNxn.225394$Jq1.170881(a)en-nntp-05.dc1.easynews.com>:

>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message
>news:5a8fs5tou6rga4q8iaela0htcdqio2grm6(a)4ax.com...
>> That's nonsense.
>
>To a large extent, yes, but I think it depends greatly on how much effort one
>goes to in destroying the media -- a disk that's just been broken into, e.g.,
>a half-dozen pieces is probably well worth putting back together.
>
>Sending the platters through a chipper should be pretty effective, I expect.

Belt sander might work fairly well.
From: JosephKK on 21 Apr 2010 23:22
On Wed, 21 Apr 2010 08:41:40 -0700 (PDT), GreenXenon
<glucegen1x(a)gmail.com> wrote:

>On Apr 21, 4:47 am, "JosephKK"<quiettechb...(a)yahoo.com> wrote:
>> On Thu, 15 Apr 2010 17:48:42 -0700 (PDT), GreenXenon
>>
>>
>>
>> <glucege...(a)gmail.com> wrote:
>> >On Apr 15, 5:31 pm, whit3rd <whit...(a)gmail.com> wrote:
>>
>> >> On Apr 15, 2:15 pm, GreenXenon <glucege...(a)gmail.com> wrote:
>>
>> >> > Can similar data recovery be performed on volatile RAM chips even
>> >> > after the power is offed.
>>
>> >> Similar, no.  Recovery, yes.  The volatility has a time decay constant
>> >> of a second or so, and it takes a long, temperature-dependent, delay
>> >> after power-off to thermalize the information to nonexistence.
>>
>> >If the time-decay-constant is a second, then will it take a second for
>> >the data to be completely lost when the power is offed?
>>
>> No.  About 1 second for 63% of the data to become thermalized.  It is an
>> exponential tail like a capacitor discharge, which is what is going on.
>
>
>Is it possible to build the memory chip in such a way that when the
>power is offed, it only takes 1/100 second for all the data to be
>completely lost even at the quantum level?

Not if it needs to be functional with useful refresh rates.
>
>Is there any other way to prevent this burn-in issue?

It is NOT burn-in. It is residual charge. See other posts on longer
retention times.
From: JosephKK on 21 Apr 2010 07:43
On Fri, 16 Apr 2010 12:20:05 -0700 (PDT), whit3rd <whit3rd(a)gmail.com>
wrote:

>On Apr 16, 12:39 am, Martin Brown <|||newspam...(a)nezumi.demon.co.uk>
>wrote:
>
>> Most users seem unaware that format does little more than alter the
>> drives directory table. That action is easily undone.
>
>True.
>
>> A full long format writing all zeroes is better
>> but would not really challenge a data recovery specialist.
>
>False; the usual 'long format' involves a search for bad blocks,
>and DOES erase and rewrite all data; nothing useful remains of
>the data on the disk, because MODERN disk drives don't waste
>any of the platter area.

Quite the opposite, modern drives reserve about 3% to 5% of the platter
for "live spare" blocks, and remap the sectors on the drive invisibly to
whatever OS/hardware is accessing the disk. It is older drives where the
space on the disk was actually physically mapped to the CHS address.

>Thirty-five years ago, there might have
>been
>some data residue.
>
>> After you have rewritten the drive with random data a few times a la
>> disk eraser a government intelligence agency might be able to get at
>> least some of the old data back but they would only bother trying if
>> they had very good reason to do so. Single bit errors in encrypted
>> and/or compressed data are fatal to decoding.

Not always.
>
>True, and there's an implication here: all hard drive data is
>modulated
>in a scheme like Manchester coding, RLL, GCR, eight-fourteen
>modulation... there's lots of schemes and lots of names. These are
>all
>history-dependent in some way, i.e. a ruined bit kills ALL the
>subsequent

not for 30 years

>data to the end of the block. There's also error-correction data,
>which
>assures the reader that his as-recovered data is wrong. There's
>no way that the hypothetical 'intelligence agency' can reconstruct the
>block well enough to match the error-correct code, and that means
>clobbering 'most of' the bits is good enough to delete the data.
>
>
>Even a few bits bad means the error-correcting code will make the
>result into evidence-unusable-in-court.

Your lack of solid understanding is glaring here. You are mixing varied
techniques at different levels of the read/write chain with wild abandon
here.
From: JosephKK on 21 Apr 2010 07:47
On Thu, 15 Apr 2010 17:48:42 -0700 (PDT), GreenXenon
<glucegen1x(a)gmail.com> wrote:

>On Apr 15, 5:31 pm, whit3rd <whit...(a)gmail.com> wrote:
>
>
>> On Apr 15, 2:15 pm, GreenXenon <glucege...(a)gmail.com> wrote:
>
>
>> > Can similar data recovery be performed on volatile RAM chips even
>> > after the power is offed.
>>
>
>
>> Similar, no.  Recovery, yes.  The volatility has a time decay constant
>> of a second or so, and it takes a long, temperature-dependent, delay
>> after power-off to thermalize the information to nonexistence.
>
>
>If the time-decay-constant is a second, then will it take a second for
>the data to be completely lost when the power is offed?

No. About 1 second for 63% of the data to become thermalized. It is an
exponential tail like a capacitor discharge, which is what is going on.
From: krw on 21 Apr 2010 18:24
On Wed, 21 Apr 2010 04:47:56 -0700, "JosephKK"<quiettechblue(a)yahoo.com> wrote:

>On Thu, 15 Apr 2010 17:48:42 -0700 (PDT), GreenXenon
><glucegen1x(a)gmail.com> wrote:
>
>>On Apr 15, 5:31�pm, whit3rd <whit...(a)gmail.com> wrote:
>>
>>
>>> On Apr 15, 2:15�pm, GreenXenon <glucege...(a)gmail.com> wrote:
>>
>>
>>> > Can similar data recovery be performed on volatile RAM chips even
>>> > after the power is offed.
>>>
>>
>>
>>> Similar, no. �Recovery, yes. �The volatility has a time decay constant
>>> of a second or so, and it takes a long, temperature-dependent, delay
>>> after power-off to thermalize the information to nonexistence.
>>
>>
>>If the time-decay-constant is a second, then will it take a second for
>>the data to be completely lost when the power is offed?
>
>No. About 1 second for 63% of the data to become thermalized. It is an
>exponential tail like a capacitor discharge, which is what is going on.

Not really. The capacitors aren't discharged sequentially. The charge on a
capacitor will leak 63% in one TC, sure, but they will all leak to 37% in that
same TC. If the threshold voltage is 50% +/- 20%, all the cells will be below
the threshold in just a little over a TC.
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