Associated Topics || Dr. Math Home || Search Dr. Math

### Rail Bend in Hot Weather

```Date: 10/13/2002 at 00:40:38
From: Jaysen
Subject: Rail Bend

Hi Dr. Math, my question to you is this:

A 20-ft piece of rail expands 1 in. in length during a hot spell. If
there are no expansion gaps (spaces found between rails that allow a
rail to expand without buckling), how high off the ground will the
rail rise?

I don't have the slightest idea of how or what you can do to attempt
to solve this problem. The only thing I do know is that it would cause
the rail to make an arc. Can you help me out?
```

```
Date: 10/14/2002 at 07:18:44
From: Doctor Jeremiah
Subject: Re: Rail Bend

Hi Jaysen,

The easiest way to figure this out is to draw a picture:

+ 20'1" +
+                         +
+                                           +
+-------------------------20'0"-----------------------+

The distance along the rail is 20 feet 1 inch, and the distance along
the rail is 20 feet 0 inches.

The curved rail is part of a circle, so the ends of the rail must be
on the circle.  This means that they can be at the ends of a radius:

+ 20'1" +
+                         +
+                                           +
+------------------------20'0"------------------------+
+                                               +
+                                         +
+                                   +
+                             +
R                       R
+                 +
+           +
+  a  +
+

Now we really have two shapes: a triangle and a section of a circle.
The circle has a circumerence of 2(Pi)R but the section is
20"1'/2(Pi)R percent of a whole circle. And since a whole circle is
360 degrees, the section is 360 x 20"1'/2(Pi)R degrees:

a = 360 x 20"1'/2(Pi)R degrees

The triangle has three sides, but it is symmetric so we could just use
half a tringle to get one with a right angle:

+----------20'0"/2---------+
+                       |
+                    |
+                 |
+              |
R           |
+        |
+ a/2 |
+  |
+

Now the sine of a/2 is (20'0"/2)/R, which means that:

sin(a/2) = (20'0"/2)/R
a/2 = arcsin( (20'0"/2)/R )
a = 2 x arcsin( (20'0"/2)/R )

Now we have two equations for a. Since they are both equal to a, they
must be equal to each other, so equate them and solve for R. Once you
know R, find h (from the triangle). The height the rail is off the
ground is R-h.

=== + 20'1" +
+     |                   +
+              |                            +
+-------------------|----20'0"---===------------------+
+                |             |                +
+             |             |             +
+          |             |          +
+       R             h       +
R    |             |    R
+ |             | +
|+           +|
|   +     +   |
===     +     ===

See if you can get it from here. If you get stuck, please write back.

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Triangles and Other Polygons
High School Trigonometry

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search