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Posers and Puzzles

Posers and Puzzles

  1. Standard member Mathurine
    sorozatgyilkos
    14 Mar '07 13:57 / 1 edit
    A cable, 16 metres in length, hangs between two pillars that are both 15 metres high. The ends of the cable are attached to the tops of the pillars. At its lowest point, the cable hangs 7 metres above the ground.

    How far are the two pillars apart?
  2. Standard member PBE6
    Bananarama
    14 Mar '07 15:51
    Originally posted by Mathurine
    A cable, 16 metres in length, hangs between two pillars that are both 15 metres high. The ends of the cable are attached to the tops of the pillars. At its lowest point, the cable hangs 7 metres above the ground.

    [b]How far are the two pillars apart?
    [/b]
    Cable hanging under their own weight take the shape of a catenary, which is expressed mathematically as the hyperbolic cosine function:

    f(x) = a*cosh(x/a) = (a/2)*(e^(x/a) + e^(-x/a))

    We centre the low point of the catenary over the origin (x = 0), at a height of 7. The pillars lie at x = -(d/2) and x = (d/2). Subbing in (0,7), we get:

    7 = (a/2)*(e^0 + e^0)

    7 = a

    Now, subbing in (d/2, 15), we get:

    15 = (7/2)*(e^(d/14) + e^(-d/14))

    (30/7) = e^(d/14) + e^(-d/14)

    e^(d/14)*(30/7) = e^(2d/14) + e^(0)

    e^(2d/14) - (30/7)*e^(d/14) + 1 = 0

    e^(d/14) = (30/7) +/- (SQRT((30/7)^2 - 4))/2 = k

    d/14 = ln(k)

    d = 14*ln(k) = +/- 19.541 (approx.)

    "d" cannot be negative (although in this case it doesn't matter since the negative value gives an answer identical to the positive value), so the distance between the pillars must be +19.541 (approx.). However, since the cable is only 16 metres long, it is not possible to string it up between the two pillars. (I think the constants need some adjustment )
  3. Subscriber sonhouse
    Fast and Curious
    15 Mar '07 01:44
    Originally posted by PBE6
    Cable hanging under their own weight take the shape of a catenary, which is expressed mathematically as the hyperbolic cosine function:

    f(x) = a*cosh(x/a) = (a/2)*(e^(x/a) + e^(-x/a))

    We centre the low point of the catenary over the origin (x = 0), at a height of 7. The pillars lie at x = -(d/2) and x = (d/2). Subbing in (0,7), we get:

    7 = (a/2)*(e^0 ...[text shortened]... ble to string it up between the two pillars. (I think the constants need some adjustment )
    Maybe it's a big rubber band cable
  4. 15 Mar '07 05:34
    I've seen one like this before, and you don't need anything other than your basic add/subtract.

    The 16 metre cable hangs down 8 metres, and hence there is only one easy-to-find solution (no special math functions needed).
  5. Standard member XanthosNZ
    Cancerous Bus Crash
    15 Mar '07 09:53
    Originally posted by geepamoogle
    I've seen one like this before, and you don't need anything other than your basic add/subtract.

    The 16 metre cable hangs down 8 metres, and hence there is only one easy-to-find solution (no special math functions needed).
    Haha, what?
  6. Standard member Palynka
    Upward Spiral
    15 Mar '07 10:08
    Originally posted by geepamoogle
    I've seen one like this before, and you don't need anything other than your basic add/subtract.

    The 16 metre cable hangs down 8 metres, and hence there is only one easy-to-find solution (no special math functions needed).
    Nice one!
  7. Standard member PBE6
    Bananarama
    15 Mar '07 14:01
    Originally posted by geepamoogle
    I've seen one like this before, and you don't need anything other than your basic add/subtract.

    The 16 metre cable hangs down 8 metres, and hence there is only one easy-to-find solution (no special math functions needed).
    Geez! Two goof-ups in a row. I was really off yesterday.
  8. 17 Mar '07 00:46 / 2 edits
    Ok got the answer and yes it is really stupid. The poles are touching hence the wire hangs down 16/2 = 8m so 7m from the ground
  9. Subscriber joe shmo On Vacation
    Strange Egg
    17 Mar '07 02:08
    Originally posted by PBE6
    Cable hanging under their own weight take the shape of a catenary, which is expressed mathematically as the hyperbolic cosine function:

    f(x) = a*cosh(x/a) = (a/2)*(e^(x/a) + e^(-x/a))

    We centre the low point of the catenary over the origin (x = 0), at a height of 7. The pillars lie at x = -(d/2) and x = (d/2). Subbing in (0,7), we get:

    7 = (a/2)*(e^0 ...[text shortened]... ble to string it up between the two pillars. (I think the constants need some adjustment )
    your hilarious
  10. Subscriber joe shmo On Vacation
    Strange Egg
    17 Mar '07 02:21
    I was reading your post in utter amazement, scrolled down and saw the answer! Wow thats some good stuff
  11. Standard member uzless
    The So Fist
    22 Mar '07 15:09 / 2 edits
    Originally posted by PBE6
    However, since the cable is only 16 metres long, it is not possible to string it up between the two pillars. (I think the constants need some adjustment )
    I think PBE6 got the correct answer since the question asks how far apart the pillars are.

    The answer, as stated by PBE6, is that it's impossible to string up a 16 metre cable "BETWEEN" 2 pillars and still have it be 7 metres above the ground at its lowest point.

    2 pillars touching each other do not allow anything to be "BETWEEN" them.