Raising Railroad Tracks

The maximum the tracks can get above the ground is $96.403$ meters with a track length of $49.592$ kilometers. Any shorter or longer track will bow at less maximum height.

This problem requires solving transcendental equations, which means it can only be done through numerical means. Refer to the figure at bottom. We have the following equations which must hold, where $a, b, r$ are unknowns, $a, b$ being angles and $r$ being the radius of the arc of the track. All linear dimensions are in meters.

$R=6371000$
$2aR+1=2br$
$RSin(a)=rSin(b)$

The maximum height $h$ for given $R$ and $a$ is then

$h=r-R+RCos(a)-rCos(b) =r-R+RCos(a)-rCos \left( ArcSin \left( \dfrac{RSin(a)}{r} \right) \right)$

but we first must need to find the value of $r$ , given $R$ and $a$ . We must solve for r the following

$2aR+1-2r ArcSin \left( \dfrac{RSin(a)}{r} \right) =0$

which we can first start with an approximation, by series expansion of $ArcSin \left( \dfrac{RSin(a)}{r} \right)$ , if we intuit that $a$ is likely to be small

$2aR+1-2r \left( \dfrac{Ra}{r} + \dfrac{ (R^3-Rr^2)a^3 }{6r^3} \right) = 0$

Solving this for $r$ gets us

$r=\dfrac{ (aR)^{\frac{3}{2}} }{\sqrt{a^3R+3}}$

Plugging this into the equation for $h$ , this time with only angle $a$ as the variable, we first find that $a$ is indeed small for maximum $h$ , approximately $0.223$ degrees. Using this number and working with the original equations directly and disregarding the approximating formula for $r$ , we come with, through further numerical approximation

$a=0.222995$ degrees
$a=0.003892$ radians
$R=6371000$ meters

$2Ra=49591.8$ meters of track at start

$b=0.668505$ degrees
$b=0.0116676$ radians
$r=2125233$ meters

$2rb=49592.8 = 49591.8 +1$ meters of track

$h=96.403$ meters