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Geeky physics comic

  1. Yes!

    82.6%
  2. No! (it sucks)

    4.3%
  3. No. (I don't get it)

    13.0%
  1. Nov 13, 2006 #1

    SF

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    I stumbled over this a couple of days ago and I found it so funny that I literally LOL-ed.
    My friends thought I was just being weird.

    Don't you think it's hilarious?
     
  2. jcsd
  3. Nov 13, 2006 #2
    That is excellent.
     
  4. Nov 13, 2006 #3

    JasonRox

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    I laughed out loud too.

    It's good.
     
  5. Nov 13, 2006 #4

    chroot

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    So Feynman liked the ugly ones, too, did he?

    - Warren
     
  6. Nov 13, 2006 #5

    JasonRox

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    Aren't there two ways to this?

    Feynman walks away with the girls, which takes them away from the guys.

    And...

    Feynman is the reason for there obsession on equilibrium, hence they miss out on the girls.
     
  7. Nov 13, 2006 #6

    selfAdjoint

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    Dearly Missed

    It's great! And if you know about Feynman, after his first wife, the love of his life, died, he turned into a ruthless bar-hopping chick exploiter. I think the comic works off that rep.
     
  8. Nov 13, 2006 #7

    Ivan Seeking

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    I thought that later he quit drinking and started playing the bongos.
     
  9. Nov 13, 2006 #8

    turbo

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    Before you can cull them, you've got to learn to cut them.
     
  10. Nov 13, 2006 #9

    Moonbear

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    Okay, now it's making sense. I didn't vote because I didn't know if I didn't get it or if it was just lame. :rofl:
     
  11. Nov 13, 2006 #10
    I know who Feynman is, and I think I have heard of Nash, but to be honest I don't get it :cry:
     
  12. Nov 13, 2006 #11
    Ok, I hate explaining jokes, but here goes.

    So the guy talks with Nash about applying Nash's theory to get the best result. They are getting technical about everything.

    Feynman, on the other hand, has always had a straight-forward approach at life. Combine that with his past with women and about just asking, etc.

    However, maybe I'm missing something. Maybe Feynman had some theory which has to do with this or something?
     
  13. Nov 13, 2006 #12
    Anything XKCD is hilarious, so much so that I have it open every morning when a new comic is up.
     
  14. Nov 14, 2006 #13

    Math Is Hard

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    *leaves with Feynman*
     
  15. Nov 14, 2006 #14
  16. Nov 14, 2006 #15

    JasonRox

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    I'm with you on that one. :!!)
     
  17. Nov 14, 2006 #16

    BobG

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  18. Nov 14, 2006 #17
    Brilliant!
     
  19. Nov 14, 2006 #18
    Actually, that second one is a pretty interesting problem. You have all the information necessary to solve it, although the solution might be a little more involved than you think.

    For example, your best strategy may not be a straight line: the raptors will supposedly run straight at you. Running along a curved path might buy you a few extra milliseconds since the raptors will have to run slightly farther as they adjust their course.

    Isn't this precisely the sort of thing they resurrected the Hamiltonian for? Solving weird problems where you know the end conditions but not the path between them? I am sure a Hamiltonian analogue exists for the wounded raptor problem.
     
  20. Nov 14, 2006 #19

    selfAdjoint

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    Yes he did, and gradually pulled himself out of his bitterness and eventually found happiness with his third wife. His second wife was a disaster, and she divorced him.

    But during the "great period" when QED was being formed, he was a chaser.
     
  21. Nov 15, 2006 #20
    I can't sleep, and my mind has chosen this one to pick on.

    It is a somewhat interesting problem, but not very in the specific case. A more general solution is probably insoluable. But some generalities might be fun to review. Either that, or my brain is seriously misfiring at this late hour (almost certainly the latter).

    In the case of identical raptor speed, sufficently low relative victim speed, the victim cannot cross the border of the triangle. His best move is to remain still. That much is obvious. We'll also assume instantly accelerating raptors smart enough to aim where the victim will be (as opposed to where he is) at this time, although it is not relevant in this case.

    With one wounded raptor, his best move is to run (possibly not at full speed) toward the wounded raptor. By symmetry, this is also a straight line.

    With equal raptor speeds and less disparate predator/prey speeds, the victim can cross the line delimiting the initial raptor triangle. His best move is full speed towards a boundary, and straight on from there. Once again, line remains straight by symmetry.

    With one wounded raptor and less disparate speeds, I am fairly sure his best path becomes elliptical some critical distance after crossing the boundary. It may be a circular segment. But I think the condition of fully anticipatory raptors is starting to break down in that analysis.

    I am fairly sure making the raptors run towards his current (as opposed to future) position yields an elliptical (or other conic section) post-critical path.

    Once you allow for finite raptor acceleration, it's starting to look a lot like the three body problem, only not quite so bad. I suspect there's still a discontinuity in the second derivative of the victim path at a critical point outside the triangle.

    Anyone care to run with this line of speculation from here (and away from the raptors)? Triple points will be taken down in evidence against you should you actually put pencil to paper and start to work out the math.

    P.S.: The raptors may have to slow at times depending on the curvature of their respective paths at that point.
     
    Last edited: Nov 15, 2006
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