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Acton's Railroad Problem

Q: In Numerical Methods that Work [1] there is a famous "railroad track problem": 1 foot of extra track is inserted into 1 mile of  railroad track, and it bows up in a circular arc. Find the maximum height it  achieves off the ground.

[Graphics:../Images/index_gr_3.gif]

How can I obtain a numerical solution to this problem and is it analytically solvable?
A: From the figure, with [Graphics:../Images/index_gr_4.gif], we obtain the following system of equations.
[Graphics:../Images/index_gr_5.gif]
Here is a one-line numerical solution.
[Graphics:../Images/index_gr_6.gif]
[Graphics:../Images/index_gr_7.gif]
Hence the maximum height it achieves off the ground is 44.5 feet.
Generalizing the equations by writing [Graphics:../Images/index_gr_8.gif] as the length of straight track and [Graphics:../Images/index_gr_9.gif] as the length of curved track, where [Graphics:../Images/index_gr_10.gif], and eliminating the nontrigonometric variables, we find
[Graphics:../Images/index_gr_11.gif]
[Graphics:../Images/index_gr_12.gif]
Even this simple equation is not analytically solvable.
[Graphics:../Images/index_gr_13.gif]
[Graphics:../Images/index_gr_14.gif]
[Graphics:../Images/index_gr_15.gif]
However, as Ronald Bruck (bruck@pacificnet.net) points out, it is easy to determine series representations for the  answer. Since [Graphics:../Images/index_gr_16.gif] and explicitly noting that [Graphics:../Images/index_gr_17.gif] depends on [Graphics:../Images/index_gr_18.gif], we obtain the series expansion for [Graphics:../Images/index_gr_19.gif].
[Graphics:../Images/index_gr_20.gif]
[Graphics:../Images/index_gr_21.gif]
Using series inversion, we obtain [Graphics:../Images/index_gr_22.gif].
[Graphics:../Images/index_gr_23.gif]
[Graphics:../Images/index_gr_24.gif]
We can use this expansion to determine [Graphics:../Images/index_gr_25.gif] and [Graphics:../Images/index_gr_26.gif]. Since [Graphics:../Images/index_gr_27.gif], the series for [Graphics:../Images/index_gr_28.gif] is
[Graphics:../Images/index_gr_29.gif]
[Graphics:../Images/index_gr_30.gif]
Since [Graphics:../Images/index_gr_31.gif], the series for [Graphics:../Images/index_gr_32.gif] is
[Graphics:../Images/index_gr_33.gif]
[Graphics:../Images/index_gr_34.gif]
Note that the order is important here--we need to determine [Graphics:../Images/index_gr_35.gif] before we determine [Graphics:../Images/index_gr_36.gif]. We truncate these series expansions as follows.
[Graphics:../Images/index_gr_37.gif]
Then, for the original problem,
[Graphics:../Images/index_gr_38.gif]
Since [Graphics:../Images/index_gr_39.gif], we find
[Graphics:../Images/index_gr_40.gif]
[Graphics:../Images/index_gr_41.gif]
This result is in good agreement with FindRoot.


Converted by Mathematica      May 8, 2000

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