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Posers and Puzzles

Posers and Puzzles

  1. Standard member Asher123
    Drunken Shogun
    18 Mar '04 22:09
    lasy joe is sitting on a beach chair, somewhere in Sudan, sipping a martini. The piece of string which surrounds the earth (which happens to be a perfect sphere) lay on the earth beside him. So joe decided to play a joke on the rest of the world so he cut the string, added a 1meter piece of string to it and then went around the world, raising the string up from the ground so that it is evenly spread around the globe again only off the ground. When he gets home, he sees his two legged cat bipod crying because his milk dish is on the other side of the string and now that it's off the ground, bipod can't get passed it.....


    here's the question: (finally) is the piece of string high enough off the ground for an average sized bipedded cat can get under it..

    the answer shocked the hell out of me
  2. Subscriber Marinkatomb
    wotagr8game
    19 Mar '04 00:20
    I'll take a guess at no, but it's obviously yes for some wierd reason. Wots the answer??
  3. 19 Mar '04 00:24
    Originally posted by Asher123
    lasy joe is sitting on a beach chair, somewhere in Sudan, sipping a martini. The piece of string which surrounds the earth (which happens to be a perfect sphere) lay on the earth beside him. So joe decided to play a joke on the rest of the world so he cut the string, added a 1meter piece of string to it and then went around the world, raising the string up ...[text shortened]... nd for an average sized bipedded cat can get under it..

    the answer shocked the hell out of me
    circumference = 2*Pi*R

    so set 2*Pi*R_old + 1 meter = 2*Pi*R_new

    then the new radius is the old radius plus 1/(2*Pi) meters.
    so the string should be about 6 inches off the ground in the final state.

    i think a cat could probably squeeze under that. It is a surprising result that the string is now that high off the ground.
  4. Standard member eyeqpc
    Robbo
    19 Mar '04 00:38
    How big is your average sized bipedded cat. You see the only bipedded cat I know of is Cheshire, and I am not sure how big C.C. is.
  5. Standard member Asher123
    Drunken Shogun
    19 Mar '04 15:00
    Originally posted by davegage
    circumference = 2*Pi*R

    so set 2*Pi*R_old + 1 meter = 2*Pi*R_new

    then the new radius is the old radius plus 1/(2*Pi) meters.
    so the string should be about 6 inches off the ground in the final state.

    i think a cat could probably squeeze under that. It is a surprising result that the string is now that high off the ground.
    correct!
    2PiR=circumference (C)
    which means R=C/2Pi
    new R=New C/2Pi
    => new R =(C+1)/2Pi
    => new R = C/2Pi + 1/2Pi
    c/2pi has already been established as R
    => new R= R + 1/2Pi meters

    so yup that's aboot 6 inches more or less. Isn't that odd?
  6. 19 Mar '04 18:25
    Certainly seems weird. If you have zero start circumference, or the size of the universe - the increase in radius when you add 1 metre to the circumference is the same.

    r = c / (2.pi)

    dr/dc = 1 / (2.pi)
  7. Standard member Asher123
    Drunken Shogun
    19 Mar '04 19:19
    exactly - 2PiR shows us more than a simple ratio it shows that the increase in circumference affects the radius at 1:1/2Pi regardless of the actual values
  8. 20 Mar '04 04:32
    Interesting.