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Posers and Puzzles

Posers and Puzzles

  1. Standard member wolfgang59
    Infidel
    06 Jul '07 06:58
    OK this is an oldie but a goldie and amazing if you have not heard it before.

    Imagine that the Earth is a perfect sphere and that we Earthlings (in a demonstration of 'brotherly love'/'Peace on Earth' etc) decide to put a nice piece of white ribbon around the equator.

    But whoops - in some mix-up with the numbers the ribbon is a metre too long! Now if this over-long ribbon were formed into a perfect circle and magicly suspended above the Earth ... roughly how far above the Earth would it hover?
  2. 06 Jul '07 17:34
    I come up with 15.9 cm, assuming that the radius of the sphere is 6380000 m.
  3. Standard member wolfgang59
    Infidel
    06 Jul '07 20:16
    Well I guess that spoils it for the non-Mathematicians! I was hoping for some intuitive guesses before the calculators came out!


    What if it were Jupiter and not the Earth though? ;-)
  4. Subscriber coquette
    Already mated
    06 Jul '07 20:35
    the same, the same as it would be if it were one meter larger around the andromeda galaxy
  5. Standard member wolfgang59
    Infidel
    06 Jul '07 21:06
    ... or a tennis ball .. and that is strange dont you think?
  6. 06 Jul '07 21:09
    ... or the whole universe ...
    ... or the pre-bangian singular point ...
  7. Subscriber AThousandYoung
    It's about respect
    06 Jul '07 23:43
    Originally posted by coquette
    the same, the same as it would be if it were one meter larger around the andromeda galaxy
    No. The larger the object, the more distance that extra 1m will be diluted over. Therefore the larger the object the lower the ribbon will be to the surface.
  8. Standard member EAPOE
    Earl of Rochester
    07 Jul '07 00:41 / 2 edits
    Originally posted by wolfgang59
    OK this is an oldie but a goldie and amazing if you have not heard it before.

    Imagine that the Earth is a perfect sphere and that we Earthlings (in a demonstration of 'brotherly love'/'Peace on Earth' etc) decide to put a nice piece of white ribbon around the equator.

    But whoops - in some mix-up with the numbers the ribbon is a metre too long! Now i ...[text shortened]... le and magicly suspended above the Earth ... roughly how far above the Earth would it hover?
    Let’s say the circumference of the earth (assuming a perfect sphere) in meters is "C" therefore using the following:

    C= PI x r

    If the ribbon is 1 meter longer the radius will therefore be

    (C+1 / PI ) - r

    There is your answer. You just need to substitute the most accurate value of “C” the earth’s circumference in meters to get your answer.
  9. 07 Jul '07 02:08 / 2 edits
    Originally posted by AThousandYoung
    No. The larger the object, the more distance that extra 1m will be diluted over. Therefore the larger the object the lower the ribbon will be to the surface.
    It's counter intuitive but that is actually incorrect. The proof, using this example and using 6380000m for the radius of the earth is this:

    6380000*2*pi + 1 = (6380000+X)*2*pi --- the LHS is the circumference of the earth plus 1 metre. The RHS is the same measurement but calculated using the new radius (6380000+X) where X is the distance above the earth the ribbon would hover.

    The rest of the proof is just rearranging the above formula:

    (6380000*2*pi + 1)/(2*pi) - 6380000 = X

    (63800000*2*pi)/(2*pi) + 1/(2*pi) - 6380000 = X

    6380000 - 6380000 + 1/(2*pi) = X --- this shows the original radius doesn't affect the distance the ribbon would be above the surface.

    1/(2*pi) = X
  10. Standard member wittywonka
    Chocolate Expert
    07 Jul '07 03:48 / 1 edit
    I'm guessing about a micrometer.
  11. Subscriber coquette
    Already mated
    07 Jul '07 04:44
    i actually had that ribbon around the birthday gift i got from my boyfriend last year and it was a real pain in the postulate
  12. 07 Jul '07 13:07
    I love how everyone keeps making wrong guesses after the right answer has been given ( 1/(2*pi) metres or 15.9cm)

    For what it's worth though when I first thought about it I also thought it would be a really tiny distance above the surface, I was surprised by how far it actually was.
  13. 07 Jul '07 14:05
    I knew this answer from the beginning, but I didn't want to interfere...

    The radius of the Earth ( or moon or nucleus of an atom or the galaxy or whatever) doesn't matter. The answer is a constant. Yes, it is counterintuitive but nevertheless a fact.

    Now another problem with the same kind of properties:
    A racing track (of some kind) has (say) 5 m in width and the race goes clockwise all the way around. How much longer is the left edge of the track compared to the right edge?

    ( "But Fabian, is this not dependant of how the track is planned? If it’s circular or if it has a lot of left and right bends?" )
  14. Standard member wolfgang59
    Infidel
    07 Jul '07 14:40
    Fabian this is even easier than my problem! I wont but any devotees to Scalextric should know the answer! (We used to work out the advantage of inside lane when Iwas a kid)

    Also any groundsmen out there who are used to marking out tracks?
  15. Subscriber AThousandYoung
    It's about respect
    08 Jul '07 03:25
    Originally posted by FabianFnas
    I knew this answer from the beginning, but I didn't want to interfere...

    The radius of the Earth ( or moon or nucleus of an atom or the galaxy or whatever) doesn't matter. The answer is a constant. Yes, it is counterintuitive but nevertheless a fact.

    Now another problem with the same kind of properties:
    A racing track (of some kind) has (say) 5 m i ...[text shortened]... of how the track is planned? If it’s circular or if it has a lot of left and right bends?" )
    2*pi*(r+5) - (2*pi*r)
    2*pi*(r+5-r)

    10*pi