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Railroad Track Problem

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RailroadProblem

Given a straight segment of track of length l, add a small segment Deltal so that the track bows into a circular arc. Find the maximum displacement d of the bowed track. The Pythagorean theorem gives

R^2=x^2+(1/2l)^2.
(1)

But R is simply x+d, so

R^2=(x+d)^2=x^2+2xd+d^2.
(2)

Solving (1) and (2) for x gives

x=(1/4l^2-d^2)/(2d).
(3)

Expressing the length of the arc in terms of the central angle,

1/2(l+Deltal)=theta(d+x)
(4)
=theta(d+(1/4l^2-d^2)/(2d))
(5)
=theta((2d^2+1/4l^2-d^2)/(2d))
(6)
=theta((d^2+1/4l^2)/(2d)).
(7)

But theta is given by

tantheta=(1/2l)/x=(1/2l(2d))/(1/4l^2-d^2)=(dl)/(1/4l^2-d^2),
(8)

so plugging theta in gives

1/2(l+Deltal)=((d^2+1/4l^2)/(2d))tan^(-1)((dl)/(1/4l^2-d^2))
(9)
d(l+Deltal)=(d^2+1/4l^2)tan^(-1)((dl)/(1/4l^2-d^2)).
(10)

This is a transcendental equation that cannot be solved exactly with a closed-form solution for d, but for l>>d,

(dl)/(1/4l^2(1-(d^2)/(4l^2)))=(4d)/l(1-(4d^2)/(l^2))^(-1) approx (4d)/l(1+(4d)/(l^2)).
(11)

Therefore,

d(l+Deltal) approx (d^2+1/4l^2){(4d)/l(1+(4d^2)/(l^2))-1/3[(4d)/l(1+(4d^2)/(l^2))]^3}
(12)
approx (d^2+1/4l^2)[(4d)/l+(16d^3)/(l^3)-1/3((4d)/l)^3(1+3(4d^2)/(l^2))].
(13)

Keeping only terms to order (d/l)^3,

dl+Deltal approx (4d^3)/l+dl+(4d^3)/l-(16)/3(d^3)/l
(14)
Deltal approx (8-(16)/3)(d^3)/l=(24-16)/3(d^3)/l=8/3(d^3)/l,
(15)

so

d^2=3/8lDeltal
(16)

and

d approx 1/2sqrt(3/2lDeltal)=1/4sqrt(6lDeltal).
(17)

If we take l=1 mile=5280 feet and Deltal= 1 foot, then d approx 44.45 feet. Solving equation (◇) numerically, we find that the true answer is d=44.498455... feet.

See also

Arc

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References

Abbott, P. "In and Out: Acton's Railroad Problem." Mathematica J. 7, 448-450, 2000.Acton, F. S. Numerical Methods That Work, 2nd printing. Washington, DC: Math. Assoc. Amer., 1990.

Referenced on Wolfram|Alpha

Railroad Track Problem

Cite this as:

Weisstein, Eric W. "Railroad Track Problem." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/RailroadTrackProblem.html

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