10

This recent question about holes dug through the Earth led me to wonder: if I wanted to dig out a tube from the north pole to the equator and build a water slide in it, which shape would be the fastest?

We're assuming a frictionless tube, of course. Let's ignore centrifugal forces. Coriolis forces do no work, and so shouldn't matter. Also, let's assume the Earth is a sphere with uniform density.

I tried to solve this problem by writing down an integral in polar coordinates for the time, then applying the Euler-Lagrange equations. However, I didn't make any progress on the resulting differential equation. Is there an analytical expression for the curve?

Community
  • 1
Mark Eichenlaub
  • 52,955
  • 15
  • 140
  • 238
  • 1
    Sounds interesting, what's the functional? – MBN Mar 22 '11 at 18:45
  • 2
    It seems like this article should be relevant: H. L. Stalford and F. E. Garrett, "Classical differential geometry solution of the brachistochrone tunnel problem", Journal of Optimization Theory and Applications, Volume 80, Number 2, 227-260, http://www.springerlink.com/content/f21724177qxptn56/ – Qmechanic Mar 22 '11 at 18:55

1 Answers1

6

Yes there is, the curve is a a hypocycloid.

See for instance:

http://en.wikipedia.org/wiki/Hypocycloid

http://demonstrations.wolfram.com/SphereWithTunnelBrachistochrone/

http://www.physics.unlv.edu/~maxham/gravitytrain.pdf

TROLLHUNTER
  • 5,172
  • 1
    More from Wolfram: http://mathworld.wolfram.com/BrachistochroneProblem.html http://mathworld.wolfram.com/SpherewithTunnel.html http://mathworld.wolfram.com/Hypocycloid.html – Qmechanic Mar 25 '11 at 21:08
  • (The first of my three Wolfram links is just for the ordinary Brachistochrone problem + friction.) – Qmechanic Mar 25 '11 at 21:30
  • If one weights the optimal trajectory with a function of the length and depth of the tunnel - i.e. real construction costs - then the best solution tends to the straight line arc. Time of travel along such arcs are constant - about 42 minutes each. Building such a network doesn't seem too unreasonable. – Andrew Palfreyman May 08 '13 at 21:32