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Posers and Puzzles

Posers and Puzzles

  1. Standard member talzamir
    Art, not a Toil
    05 Nov '13 11:47
    Simple basic geometry this time.

    A Ferrari Testarossa is 4485 mm long and 1130 mm wide. It's taken to a track and field area where the field inside the track consists of a square in the middle and two semicircles at the ends, with a total perimeter length of 400 meters. How wide does the track need to be for the car to be able to go a full lap?
  2. Standard member forkedknight
    Defend the Universe
    05 Nov '13 16:19
    Are we to assume there is a wall on the inside and outside perimeter of the track, that the car is a perfect rectangle, and that the car's turning radius is irrelevant?
  3. Standard member forkedknight
    Defend the Universe
    05 Nov '13 16:49 / 3 edits
    Definitions:

    r = radius of semi-circle at end of track
    L = 4485mm (length of car)
    w = 1130mm (width of car)
    w' = width of track
    r' = r + w'


    Calculation:

    r = 400m / (2pi + 4) ~= 38898.5mm
    theta = tan-1( (L / 2) / (r + w) )
    r' = (L / 2) / sin(theta)
    w' = r' - r

    Plugging all of that into wolfram alpha (because I'm lazy)

    w' ~= 1192.77mm

    *edit*
    For this to actually work, the car would need to be able to steer both the front and rear wheels
  4. Standard member talzamir
    Art, not a Toil
    05 Nov '13 20:26 / 3 edits
    nice work, and yes, the 400 meters is as you wrote it, and yes, I thought that the Ferrari looks like a moving rectangle when seen from above, either with four-wheel steering or with a very slippery track and no steering needed at all.

    I didn't use trigs.

    Distance from the center of the semicircle to the middle of left side of the car (if driving counter-clockwise) = R.

    Distance from the center of the semicircle to the right front / back corner = ((R+w)^2+(L/2))^(1/2)

    So the width is of the track is ((R+w)^2+(L/2))^(1/2) - R.

    which gives exactly the same result, of course.