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The old man and the bridge

  1. Sep 4, 2009 #1
    An old man has built his beautiful house on an unused bridge for excellent scenic views of the landscape. However there is a problem, demolition crew has arrived and have set charges to the sides of the bridge that connect it to the land-masses. The demolition crew does not know of the old man and vice versa, because the bridge is very long and the old man has resided in the middle of it. Meanwhile, the old man while enjoying the natural scenic vistas is about to set all possible positions on a 3x3x3 Rubik's Cube, he has clumsy hands and can only do one position per second. He is daring to do this because he has stopped aging, and thus he can live forever, provided nothing physical happens, such as the destruction of the bridge. The bridge measures in height of 20,000 feet and in length an impressive 1.9x10^15 km. The bridge is only linked on either sides of land-masses and is not a suspension bridge. There is standard atmospheric pressure and gravity, just like on Earth. At 20,000 feet, the old man's house is pressurized. Also, he has several massive stockpiles of food that will last for almost unlimited amount of time. He spends 9 hours on sleep, 3 hours on food and exercise, and from now on will be spending 12 hours on Rubik's Cube in a single 24 hour day, everyday. Sadly, just as he makes his first position on the Rubik's Cube, the charges are detonated.

    How many position(s) of the Rubik's Cube was the old man able to do before he died?
     
  2. jcsd
  3. Sep 11, 2009 #2
    Funny, for me to understand this brain teaser is a brainteaser itself :)
     
  4. Sep 11, 2009 #3

    DaveC426913

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    Shoot. I thought I'd found some wiggle room.

    A bridge 1.9x10^15 km wrapped around the Earth would be gravitationally stable, floating without need of support. So, the answer would be that he can finish the cube.

    But you did say "...just like on Earth...", so it's not Earth.

    Clearly, the old man is living on the outer surface of a galactic Dyson Sphere.
     
  5. Sep 11, 2009 #4
    Yeah, I'm not sure I understand what's being asked. For instance, it says "he can live forever, provided nothing physical happens, such as the destruction of the bridge." And we know that the detonators are set off precisely as he completes 1 "position" of the cube. You would think that setting off the detonators is "something physical". He's in free-fall when the bridge starts falling, so he's going to find it hard to exercise and eat and sleep normally. So it would seem by that logic that the answer is 1.

    But that's obviously just poor wording. It seems there's something else that's supposed to be going on, given the overly complex statement of the problem. I would have assumed that was merely how long it would take for the bridge to hit the ground, but then the length of the bridge would be irrelevant. So by stating that fact, it's as if the length of the bridge is somehow implying that the center of the bridge won't start falling immediately when the two ends are cut (sounds like something out of Looney Tunes). And given that it's probably NOT on Earth, the implication is that the bridge doesn't follow the curvature of the ground (if it exists), but that instead the bridge and ground are perfectly flat in the area in question, with equal gravity and atmospheric conditions throughout.

    So, as stated, the answer to the question is 1. With clarification, the problem would probably still be very simple, but unnecessarily complicated.

    DaveE
     
  6. Sep 11, 2009 #5
    I think the OP wants us to assume that the point at which the bridge starts to crumble moves toward the old man at the speed of light. However, real bridges don't crumble so efficiently (real bridges aren't quite so long either, perhaps this bridge is an efficient crumbler). Since the midway point of the bridge is 9.5 x 10^14 kilometers from the ends, and the speed of light is 3 x 10^6 kilometers per second, it will take just over 316666666 seconds for the signal to reach him and he can put the cube in that many positions. However, the crumble point will not move at anywhere near that speed and he will have time for many times that many positions. In the meantime, the old man can look at us in the same way that Wile E. Coyote does after the bridge falls away beneath his feet, and he is still suspended in air. Does anyone know how to estimate the speed of the crumble point for real bridges?
     
  7. Sep 11, 2009 #6
    Ahhh, that would make some sense, I suppose. So the missing parts of the equation would be how fast the "falling" ripple can progress through the bridge, and the terminal velocity of the bridge in free-fall (since the implication of the bridge's height even being mentioned is that he'll have the free-fall time to continue working).

    DaveE
     
  8. Sep 11, 2009 #7
    And to rule out the possiblity of him simply stepping off the bridge and walking away.
     
  9. Sep 13, 2009 #8
    Teaser Visualized

    far7u8.png

    *Not to scale.
     
  10. Sep 13, 2009 #9
    Re: Teaser Visualized

    This image is pretty and all, but I'm not sure what it's meant to clarify.

    1) How many times does the bridge wrap around the "Hyper Planet"?

    2) Are we to assume the bridge is straight in the "X" and "Z" directions (as per your graphic), but is NOT straight on the "Y" axis?

    3) How fast can the ripple of "falling" pass through the material contained within the bridge?

    4) What's the terminal velocity of the bridge when falling?

    5) How far from the center of gravity of the hyper planet is the bridge?

    6) Does it matter that the hyper planet has a "crust" and is "hollow"?

    7) What specifically is the "anything physical" that stops the old man from doing any more
    positions on his Rubik's cube?

    8) How long is a day on Hyper Planet? Is it relevant that he spends 9 hours for sleep and 3 hours for exercise, rather than simply to state that he spends 12 hours per day on the Rubik's Cube?

    9) It's unlikely to assume that the different positions of the cube are attainable linearly. For that matter, I'm not sure that all theoretical possible orientations are attainable through the transformations provided by the cube. Are we expected to assume that all positions are attainable, or merely the possible ones? And are we expected to assume that all positions are attainable linearly? That is, for every second he spends, he can ALWAYS get to a distinct never-before-done position on his cube within 1 second of attaining the last position.

    DaveE
     
  11. Sep 13, 2009 #10
    The bridge is far too strong to crumble so it falls in its entirety at once. When separated from the cliff, the lack of force must be propagated through the entire bridge before it can fall. You can ignore the time it takes to fall 20,000 feet. The height was only given to make sure that the bridge doesn't interact with the planet as a binary system, due to its thickness. The old man never had the time to escape. The reason why I made this riddle was to highlight the point that special relativity can cause things as weird as keeping a bridge standing in the air for centuries without any support, as if floating. And with that, many members of the general public can not accept or even phantom such a phenomenon, they just deny and are quick to call Einstein wrong. It is sad that people can not realize that there is more than just their intuitions.

    Now on the impressive length. The bridge is so long that it would almost stretch from the Earth to the neutron star, RX J185635-3754 http://www.hubblesite.org/newscenter/archive/releases/2000/35/".
     
    Last edited by a moderator: Apr 24, 2017
  12. Sep 13, 2009 #11
    Re: Teaser Visualized

    The bridge only occupies a small fraction of the Hyper Planet, so it does not wrap around. The Hyper Planet has a diameter of 20,000 light-years, compare to 14,000 light-years diameter of the Large Magellanic Cloud, galaxy, http://en.wikipedia.org/wiki/Large_Magellanic_Cloud" [Broken].
    Same as on Earth.
    The crust provides the gravity continuously across the bridge. The crust has the thickness of almost equivalent to the diameter of Earth.
    Absolutely, or otherwise the planet will turn into an hyper-ultra-super massive black hole with a mass more than that of the entire Milky Way galaxy, and thus becoming the most massive black hole in the universe.
    Its days are modeled after Earth. So 24-hour-artificial days, controlled by massive lights.
     
    Last edited by a moderator: May 4, 2017
  13. Sep 13, 2009 #12
    The old man is quite advanced, LOL.
     
  14. Sep 13, 2009 #13
    Yes, that was the main purpose. :biggrin:
     
  15. Sep 13, 2009 #14

    DaveC426913

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    In case it does not go without saying, this is not a riddle. There's no trick; it's simply a math problem.

    Paring the problem down slightly:

    How many seconds are there in 1.9x10^15km * c * 0.5 ?
     
  16. Sep 14, 2009 #15
    Rather: How many seconds are there in 1.9x10^15km * 0.5 / c?
    As archis said, the brain teaser part is trying to figure out what the OP means.
     
  17. Sep 14, 2009 #16

    DaveC426913

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    Meh. Multiply. Divide. Picky picky. :biggrin:
     
  18. Sep 14, 2009 #17
    Rather: How many seconds are there in 1.9x10^15km * 0.5 / c?
    As archis said, the brain teaser part is trying to figure out what the OP means. The idea of a bridge suspended in air because it is too rigid to break up, and too long to get the signal to fall, is too unphysical to interest me. For 50 cents more you could have made the bridge out of antigravity material and it would never fall. I like this puzzler better:

    Who hears a singer first, the audience in the music hall where the singer is, or the audience listening to their radios at home a thousand miles away.
     
  19. Sep 14, 2009 #18

    DaveC426913

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    Double post.

    Assuming the microphones are right on the stage (i.e. closer than the audience), and that there's no post-processing of the recording, it is quite possible that the home-listeners hear it first.
     
  20. Sep 14, 2009 #19
    Yeah, and unnecessarily complex. There's also the extra bit that he's only spending half of his time doing the cube. So divide your result R by 86400, round down to get the number of days he's spent, and then multiply that back by 86400 for the total seconds S. Then calculate R modulo 86400, and if it's less than 43200, add it to S, otherwise, just add 43200.

    DaveE
     
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