h a l f b a k e r y"It would work, if you can find alternatives to each of the steps involved in this process."
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The idea is that this is a truly halfbaked invitation to a
halfcon, in Brisbane, Australia.
Heads up for anyone in the Brisbane area on Tuesday 19
June, 2012. There will be a halfcon in the lobby at the
Brisbane Hilton Hotel.
Of course, this time may have already passed, in your part
of
the world, so you can think of it as a Heisenberg Halfcon, if
that is the case.
[link]
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Ooooh, oooh! I think I can just make that one. You will recognize me as I won't be wearing a carnation in my lapel. [+] |
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Apologies, I won't be able to have made it. |
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I was there tomorrow, where were you guys? |
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They would have been being happy at the likely
participation of their online friends. |
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You'll be laughing on the other side of your face in around six
centuries' time when the Arabs have taken over Australia and
there really is a halfcon there. I have no explanation for their
adoption of Gregorian month names though. |
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I plan to be dead in six centuries from now. |
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//this time may have already passed, in your part of
the world// |
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Nah. When it's 2012 in the UK, it'll still only be about
1957 in oz. |
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I'll make an effort to send some nutrinos that passed briefly through me in that direction. If I'm really extraordinarily lucky I might deflect one. |
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I'll also make an effort to pop some particles in and out of existence there that may or may not have any association with my person whatsoever. |
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Ah, you poor insubstantial and tiny person who is not
a neutronium sphere a parsec in diameter,
[RayfordSteele]. |
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Will Schrödinger be there? |
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//'ll make an effort to send some nutrinos that passed briefly through me in that direction.// |
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Hmmm.
Apparently the primordial neutrino flux is 3*10^12 (neutrinos per square centimeter per second.
Or 3*10^16 per square meter per second.
Presumably these are evenly distributed.
Consider a sphere with the surface area (in m^2) of the neutrino flux (per sec per m^2). Each square meter of the surface will pass a neutrino from that central square meter on average once a second.
Therefore, the range at which you can reliably emit a neutrino from a 1m^2 'source' to a 1m^2 target in a few seconds is:
sqrt(3*10^16/(4*pi))=2.4*10^15 meters |
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UnaBubba, I'm sending you more neutrinos than you'll ever need. Use them wisely. |
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// UnaBubba, I'm sending you more neutrinos than you'll ever need. Use them wisely // |
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We're sending you a cat, a detector and a radioactive source in a sealed box. Second class. Via Bulgaria, Tristan da Cunha and Solihull. Don't open the box, it will probably smell baaaaaad ... but you won't know, unless you open the box. |
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_possibly_ fixed. Think I neglected a bracket in the written formula, so I put one in. I might have made other mistakes in the calculation (it took longer than I had to write the anno, so I was in a hurry). I've still not checked it properly though - hope to get back to it. |
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Do people from here meet up for halfbakedness or was this a one-off? |
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Usually we just meet in small groups and laugh about
those not in on the joke, [Phront]. |
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I can see why there aren't many. |
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There's a certain satisfaction in seeing someone prove me right :) |
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I wasn't there again, today. |
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