A bit OT, but diy

[Dave]

I do a few minor repairs on cameras and phones that have got wet, e.g. dropped into a loo or streams and other things that can easily be fixed.

I have been handed a camera memory stick to look at. I was told that it was dropped and was now not working.

Is this possible? I would have thought that it was light enough not to hit the floor with very much G.

Neither of the 2 computers here recognizes that it is in the card reader (2 readers, one on each computer) I have cleaned up the contacts, but it is still not readable.

Any ideas, anyone?

Dave

[ecacguard-goog]

Not happened to me before with camera memory sticks, but it's certainly possible (and I've had it with the usb drives).

In any case, there's not a great deal you can do - if you've cleaned the contacts, and tried it in two PCs and it's not even recognised as attached (let alone readable) you're basically stuffed. With camera memory, there's not really anything significant in the way of casing or similar you could open up and fiddle with, even if you were very handy with a soldering iron and gaffer tape :-)

You could try it in a camera too but not much else to suggest. If it's recognised as attached but not readable, try reformatting the stick either on a camera or a PC.

If you get no joy, well at least these things are getting cheaper and cheaper.

Ed

[Dave]

That was my immediate thought

:-) Agreed

Owner has tried it in his camera, so I have come to the conclusion the the stick is dead. As you say, they are so cheap, that they can be bought for a few quid. Mind you, I did advise him to buy smaller memory cards and not to put all his eggs in one basket again. He had photos of his very young daughters on the card that he has just lost.

Dave

Many thanks for your time

Dave

[John Rumm]

It would still be possible to break an interconnect wire between the connector and the flash. It only takes a lifted track or pin on the connector etc.

Depends on the value of the data. Taking apart and retouching all the soldered connections may solve it, but on small surface mount devices this is not an easy task, and best left to someone who has experience of doing this and the required equipment. Another possibility would be a professional data recovery company. If the flash device is undamaged then chances are that it still has a record of the data and it is just access that is impaired.

[Dave]

True.

I have all that, but the cost to time ratio does not add up. Better to bin the card and live and learn from the experience. (eggs in one basket comes to mind)

Once again the cost to time ratio kicks in. I have lots of specialist software that can recover the photos, but only after I can read the card.

I think I will advise him to try formatting the card and see what happens. Total loss will be less than £20-00 at the most, if the card is dead. Compare this to the cost of data recovery and it does not add up.

Dave

[TD]

What make was it? If it was a recognisable make like SanDisk or Kingston I'd get in touch with the company to see if they can do anything for you. Nothing to lose plus the chances are it will still be under warranty.

I've had 2 cards fail on me. One compact flash fell from pocket height on to a lino floor. It was a no make PQI. It wasn't recognised in any device and after asking about I put it down to a stress fracture. It was exchanged no problem. The other was a SanDisk memory stick that got corrupted in a card reader. Regardless of endless customer support from SanDisk it still couldn't be recognised in any device. Again it was exchanged no problem.

I certainly no longer think these flash memory devices are indestructible.

T.

[Adrian C]

Don't tell him to reformat the card!!! If the pictures are irreplacable then they are worth a lot more than the material cost of the card, which is now peanuts (check Amazon, Ebuyer).

There are utilities such as Photorec that will work at a low level ignoring file system problems, to recover images.

There are also companies that may be able to help.

Like

[Cicero]

================================== It might be worth trying an hour or so in the 'fridge to try and recover the data. This sometimes works with ailing hard drives. I know it's different technology and probably won't work but you've got nothing to lose by trying.

Cic.

[TD]

How will he be able to reformat the card if nothing recognises it? If it can be reformatted (recognised) then it should be able to get the data back.

[Dave]

You could well be right here. Owner is an Afghan refuge. He has been over here for well over 6 years and has brought his wife over and raised a family. Good luck to him.

The card is branded Sony, but I do not know when he purchased it

Many thanks for that. I am going to print this out for is information

Dave

[Dave]

Whoops :-(

Noted

Dave

[Dave]

Worth a punt :-)

Thanks

Dave

[robgraham]

I've used Zero Assumption Recovery successfully to retrieve seemingly lost data on a camera card.

Rob

[Andy Burns]

Being light means less force, but it doesn't reduce gravity ;-)

[Steve Walker]

If it's light enough compared to its surface area then air resistance will make it fall slower. Hitting the floor slower will reduce the rate of decelleration that it experiences, which is of course frequently measured as multiples of g. That's what I understood Dave's post to refer to.

Just to be really pedantic, the memory card of course also possesses a miniscule force of gravity and the lighter it is, the less gravity it has!

SteveW

[Bob Eager]

Following the pedantic line, that would of course be "minuscule"...!

[Steve Walker]

You're right. One question though. Were you just correcting my spelling or were you cleverly using minuscule in its usage as referring to lower case letters where I'd used g rather than G too?

SteveW

[Bob Eager]

No, I wasn't being that clever! But a good point!

I just couldn't resist the spelling bit, since the word 'pedantic' had been used. At least you didn't say you were a 'pendant', which seems reasonably common!

[OG]

Aren't they the people that just hang about waiting for others to make mistakes?

[John Rumm]

g would be acceleration due to gravity, and G would be the universal gravitational constant - as in g = (G x m1 x m2) / d^2