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The Buckled Rail
Submitted by Miran from Louisville, Kentucky, 1/18/2000.
Original answer by Esther Fontova; this article by Allen Stenger.
A railroad one mile long is anchored at both ends. On a hot day it expands
one foot and buckles. Approximately how high off the ground is it at its
midpoint?
[Editor's Notes:
(1) You can assume the buckled rail has the shape of an
arc of a circle. Books on Strength of Materials
would assume a parabolic shape.
(2) The problem asks for an "approximate" answer;
how "approximate" is your answer? Try to get upper and lower bounds for the true height.]
Hint 1
Draw a picture. Include the radii of the arc.
Hint 2
Here's a picture (not to scale). The two arcs marked
c
are the buckled rail, and the two line segments marked
d
are the straight line connecting the ends of the rail and lie on the ground.
We've used letters here for all the measurements; we already know that
2c = 5281 feet and 2d = 5280 feet.
If we knew the length k we could calculate the
height h using the Pythagorean theorem, but we don't
know k. Try to get an upper bound for k
(this is easy),
then get an upper bound for h.
Then try to get a lower bound for k (this is a lot
harder) and use that to get a lower bound for h.
Hint 3
The easy upper bound for k is c
(k is a straight line, so its length is less than
any other curve between the same endpoints).
Therefore we get an upper bound for h as
h=\sqrt{k^2-d^2} < \sqrt{c^2-d^2} = \sqrt{2640.5^2-2640^2} = \sqrt{2640.25} < 51.39 \mathrm{\ ft}.
This seems much too big; imagine that a mere increase in length of
1 foot over a 1 mile stretch would raise the midpoint 51 feet!
Of course, our figure is an upper bound, so the true
value might be a lot smaller. We'll investigate a lower bound next.
The next three hints give a way to get a lower bound for
k.
Hint 4
Show that
k > (d + c)/2
and use this to get the lower bound for h.
Hint 5
Write the lengths in terms of r and the angle \alpha.
We can get these from the definitions of the sine, of the angle (in terms of
arc length), and by bisecting the angle \alpha
to get a right triangle with hypotenuse
r
and one side
k/2.
We get
\begin{eqnarray}
d &=& r \sin \alpha \\
c &=& r \alpha \\
\frac{k}{2} &=& r \sin \frac{\alpha}{2},
\end{eqnarray}
so that we want to show
2 r \sin \frac{\alpha}{2} > \frac{r \sin \alpha + r \alpha}{2};
or, writing \beta=\alpha/2,
\sin \beta > \frac{\sin 2\beta}{4} + \frac{\beta}{2}.
Hint 6
Draw a sector of the unit circle.
We will prove the inequality by using the three areas in the figure:
Figures
A and A+B
are triangles with height
\sin \beta
and bases
\cos \beta
and 1 respectively, and
A+B+C
is a sector of the unit circle with angle
\beta.
Therefore the areas are (using area = (1/2) base times height or area = (1/2) arc length):
\begin{eqnarray}
A &=& \frac{1}{2} \sin \beta \cos \beta = \frac{1}{4} \sin 2\beta \\
A+B &=& \frac{1}{2} \sin \beta \\
A+B+C &=& \frac{\beta}{2}.
\end{eqnarray}
From the figure, the area B is half the area of the rectangle abcd,
and the sliver C is part of the other half and so B>C.
The equations above then give us the inequality
\left( \frac{1}{2} \sin \beta - \frac{1}{4} \sin 2\beta \right) > \left(\frac{\beta}{2} - \frac{1}{2} \sin \beta \right)
and therefore
\sin \beta > \frac{\sin 2\beta}{4} + \frac{\beta}{2},
which is what we wanted to prove.
Now we can find a lower bound for h.
Click here for rest of the solution.
The Rest of the Solution
The lower bound is
k > \frac{d+c}{2} = \frac{2640+2640.5}{2} = 2640.25
h = \sqrt{k^2 - d^2} > \sqrt{2640.25^2 - 2640^2} = \sqrt{1320.0625} > 36.33 \mathrm{\ ft}.
So we know the true height at the midpoint really is somewhere between
36.33 feet and 51.39 feet, all from a little 1-foot increase in the
total length!
Observe how sensitive the answer is to little uncertainties in
the length k; we know k is between
2640.25 and 2640.5, a tolerance of 1/4 foot, but this causes
an uncertainty in the height of about 15 feet!
A More Precise Estimate
Another way to work this problem is to figure out
the angle \alpha.
By dividing the first two equations in Hint 5 we cancel the
unknown r and we get an equation
\frac{\sin \alpha}{\alpha} = \frac{2640}{2640.5}.
Unfortunately there is no explicit solution for this equation,
so you have to get a numeric approximation.
You can do this with calculus, or you can solve it with a graphing calculator
such as the TI-83 Plus.
Then you can figure
r and then h.
The reference by Acton works this out in
detail, and gets \alpha=0.033708 radians and
h=44.499 feet.
References
- Forman S. Acton, Numerical Methods That Work,
Mathematical Association of America, 1990, pp. 3-4, 67-69.
Works out a more precise estimate.
-
Weisstein, Eric W. "Railroad Track Problem." From MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/RailroadTrackProblem.html
An online copy of Acton's solution.
- "The Buckled Rail: Three Formulations" by James E. Mann, Jr.;
College Mathematics Journal, v. 29 no. 2 (March 1998), pp. 138-141.
This article considers three different models for the shape
of the buckled rail: triangular, circular (our model), and sinusoidal.
- Click
here to view the original problem submitted by Miran.
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