Track and Field Car

talzamir
Posers and Puzzles 05 Nov '13 11:47
1. 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. 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. forkedknight
Defend the Universe
05 Nov '13 16:493 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. talzamir
Art, not a Toil
05 Nov '13 20:263 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.