I've a compact flash storage card. When will it be heavier, when I fill all bits with zeros or ones?


Let the games begin... :P

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The zeroes of course. With a 1, you get the weight of the digit. With a 0, you get the weight of the digit along with the weight of the hole in the middle.

Neither. When you put a bit of data on a card you are just changing the orientation of the bit that is already there. You aren't adding anything but organization to the card.

Kizarvexis

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That would depend on which way you hold the card with respect to the local geomagnetic field...

the zeros:

if you have zeros, you might as well have a hundred or a thousand times zeros, that still remains zero. So for every 1, you can take as many zeros as you want! So if you load you card with zeros, and it feels like putting a lot of redundancy in it, you might no longer be able to lift your PDA!!

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This was my instinctive guess, too (that it was a magnetic switch, so the heavier one would be whichever was the higher energy state, depending on its alignment with any external fields). However, after a quick review of just how a flash memory card works, I see that there is a thin oxide layer for each cell. This layer is uncharged to store a 1 and given a negative charge to store a 0. This means that there are some extra electrons in the material when storing 0, so the mass should be a little higher.

Well, As I remember it, a Flash memory device is in essence just an array of field effect transistors with a capacitor doped on top of each(actually it is a bit more involved, but it is the capacitors that matter). I think a low cell voltage is counted a "1" and you charge the capacitor to set the bit to "0". So, the question is, is a charged capacitor heavier than a discharged one...

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RAM uses capacitors, but since capacitors leak over time, RAM loses stored information if it's not constantly powered. Flash memory is a little more complex so that it can hold datd when unpowered, but you're at least correct that there's an induced charge (on an oxide layer between a floating gate and a control gate in this case) to change the setting from 1 to 0. Whether charging a neutral object increases or decreases mass depends on whether you charge it negatively (adding electrons) or positively (removing electrons). Since in this case it's a negative charge, that means the mass increases (the extra mass of all those electrons).

That reminds me of the recent "video compression" explanation in Dilbert.

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0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 ...

OMG!

You didn't think he was serious did you?

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Couldn't you check the effect of a magnetic field by using a strong magnet to pick up a blank card vs. one filled with data and see which is more strongly attracted?

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Well, you never can be quite sure on a message board, especially this one, but I thought he might be kidding.

In any case, some other posts are tending toward the idea that there might be a very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, slight difference between a 1 and a 0 on an atomic level.

Kizarvexis

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"We don't just borrow words; on occasion, English has pursued other languages down alleyways to beat them unconscious and rifle their pockets for new vocabulary." - James D. Nicoll

Don't be silly - 0s and 1s weigh exactly the same.

However, 0's are rounder than 1's, so if there are any kinks in your computer cables they will be able to get round the bend but the 1's might get stuck. Keep your cables nice and straight.

Experimentation has shown that while all 1's, a flash card will sink when placed in water.

If all 0's, it will in fact float. . . just as a large rock will sink, while in fact a small pebble will float.

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That would imply you're giving the whole card an overall charge, which seems unlikley to me. Surely we're talking about displacing charge around/in/either-side-of the oxide layer, with the overall total remaining at zero.

Therefore, they weight the same 8)

Well, you only need one place to get a zero. For a 1, on the other hand, you have to enter 0,999999999999999... so that's definitely heavier. On the other hand, a zero has two signs, and a 1 only 0,99999999999..., so that might compensate a bit (pun intended!).

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If everyone keeps on formatting flashes into zeros and never use them again, will we eventually run out of electrons?

I ran out of electrons once last year. Oddly though it was a positve experience overall.

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only if it weighs less than a duck

If it weighs more than a duck it is a witch and we´ll have to burn it at the stake

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The charge comes by connecting it to an external voltage (the power source for the memory) briefly, so I think that it does actually induce a net charge, though it would obviously be quite small. Since it's fun to seriously investigate something completely trivial, I'll see whether I can find out more detail. Of course, even if the extra electrons do come from an external source, that would mean that the batteries in your digital camera become lighter by exactly the same amount, so it would still all balance out, sort of.