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May 22, 2012 at 7:34 comment added abdullah kahraman @stevenvh ah, that is exactly what I need, thanks!
May 22, 2012 at 6:37 comment added Connor Wolf @stevenvh - At that point, I think you can let the heat-death of the universe boil them for you.
May 22, 2012 at 6:36 comment added stevenvh @abdullah - 128 bits give you a maximum delay of \$~\approx 5.4 \times 10^{23}\$ years. That should be enough to boil those eggs.
May 22, 2012 at 6:35 comment added Connor Wolf Frankly, the inaccuracy in the length of a year is only a ~0.06% error. I would expect the oscillator drift to be orders of magnitude more then that, and that's not exactly something I can fix by increasing the precision of the constants I am using.
May 22, 2012 at 6:33 comment added Connor Wolf @stevenvh - I added a note regarding you point about the length of a year. It's right before the comment Lastly, this whole exercise is rather silly.
May 22, 2012 at 6:33 comment added stevenvh "Really, it's just a (somewhat) silly exercise". Of course, I agree on that. But even with the\$~\approx\$ I expect that all but perhaps the last digit are correct.
May 22, 2012 at 6:33 history edited Connor Wolf CC BY-SA 3.0
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May 22, 2012 at 6:30 comment added Connor Wolf Really, it's just a (somewhat) silly exercise, showing how very little RAM is required to achieve ridiculously long delays.
May 22, 2012 at 6:28 comment added Connor Wolf Really, it's more that truncating accuracies in a modern digital calculator is harder then just using the numerically precise, but realistically inaccurate value. I did use \$\approx\$ (the TeX "approximate" symbol).
May 22, 2012 at 6:27 comment added Connor Wolf @stevenvh - It's what my calculator output? I certainly don't claim it's that accurate, hell, 0.1% drift in the system oscillator will render such long delays highly imprecise, let alone the drift in a RC oscillator (if such is used).
May 22, 2012 at 6:27 comment added stevenvh The length of the tropical year is about 365.24219 days. Your calculation for the number of years starting with 15 digits for the number of seconds has an error in the 4th digit. Just to illustrate how pointless it all is.
May 22, 2012 at 6:11 comment added stevenvh Representing your result in 15 significant digits is ridiculous. Even that is not the exact value, so what exactly is the point? Many people don't seem to realize how little \$1\$ in \$10^{15}\$ is. Even in the same line you drop 11 of the digits, making a rounding error for 37.06 in the process. So you start with 15 digits, and end up with 4 where the last one is wrong...
May 22, 2012 at 6:07 comment added abdullah kahraman What if it is not enough?
May 22, 2012 at 3:46 history edited Connor Wolf CC BY-SA 3.0
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May 21, 2012 at 20:59 history edited Connor Wolf CC BY-SA 3.0
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May 21, 2012 at 20:51 history answered Connor Wolf CC BY-SA 3.0