4 ms·

1. Convert all the data to decimal (digits 0-9). 2. Put a decimal point in front of this long string. The result will be a rational number between 0 and 1.

by edw519 16y ago

1. Convert all the data to decimal (digits 0-9). 2. Put a decimal point in front of this long string. The result will be a rational number between 0 and 1. Call it x. 3. Get a titanium rod exactly 12 inches long. 4. Using a fine laser, etch a line in the rod precisely 12x inches from the end. 5. Done. Precise, durable, elegant, compact, and green. EDIT: </sarcasm>

bayes 16y ago
It's certainly possible to imagine universes in which that would work. But in ours, where stuff is made of atoms, I can't see you positioning the mark on the rod any more precisely than the width of an atom, which I think is about 10 to the -10 meters. So I'm guessing you could only encode 30 or 40 bits, even with super-advanced etching and measuring equipment.
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JadeNB 16y ago
Ah, you added the `</sarcasm>` tag while I was responding …. Anyway, it was an excuse to break out Frink (http://futureboy.us/fsp/frink.fsp http://futureboy.us/fsp/frink.fsp), which reports that the resolution required, which (I think) is (1 foot)/(50 terabytes/byte), is 6 * 10^(-15) m, i.e., on the order of the diametre of a proton. Honestly, I thought it would be much smaller. Another problem is that any rod etched in this way will have two decodings. :-)
- JadeNB_redux 16y ago
  Gah, anti-procrastination settings prevented me from correcting this, but it's too stupid to let stand. At least this post demonstrates once again the important principle that a smart (computer) calculator is no match for a dumb (human) calculator. Of course, n bytes can store a number up to 2^(8n), not up to n. Thus, the number we'd be recording has order of magnitude (50 terabytes/bit)×log[2]/log[10] ≈ 1.20e+14, so we'd need to distinguish that many digits after the decimal point. Conversely, as bayes pointed out (http://news.ycombinator.com/item?id=1585498 http://news.ycombinator.com/item?id=1585498), if we have a resolution of 10 digits after the decimal point, then we can handle about 10*log[10]/log[2] ≈ 33 bits. (EDIT: To be clear, it's the same person posting; I just can't stand to wait 154 minutes to correct the error.)

by edw519 16y ago

bayes 16y ago

It's certainly possible to imagine universes in which that would work. But in ours, where stuff is made of atoms, I can't see you positioning the mark on the rod any more precisely than the width of an atom, which I think is about 10 to the -10 meters. So I'm guessing you could only encode 30 or 40 bits, even with super-advanced etching and measuring equipment.

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JadeNB 16y ago

Ah, you added the `</sarcasm>` tag while I was responding …. Anyway, it was an excuse to break out Frink (http://futureboy.us/fsp/frink.fsp http://futureboy.us/fsp/frink.fsp), which reports that the resolution required, which (I think) is (1 foot)/(50 terabytes/byte), is 6 * 10^(-15) m, i.e., on the order of the diametre of a proton. Honestly, I thought it would be much smaller. Another problem is that any rod etched in this way will have two decodings. :-)

JadeNB_redux 16y ago
Gah, anti-procrastination settings prevented me from correcting this, but it's too stupid to let stand. At least this post demonstrates once again the important principle that a smart (computer) calculator is no match for a dumb (human) calculator. Of course, n bytes can store a number up to 2^(8n), not up to n. Thus, the number we'd be recording has order of magnitude (50 terabytes/bit)×log[2]/log[10] ≈ 1.20e+14, so we'd need to distinguish that many digits after the decimal point. Conversely, as bayes pointed out (http://news.ycombinator.com/item?id=1585498 http://news.ycombinator.com/item?id=1585498), if we have a resolution of 10 digits after the decimal point, then we can handle about 10*log[10]/log[2] ≈ 33 bits. (EDIT: To be clear, it's the same person posting; I just can't stand to wait 154 minutes to correct the error.)