railroad track expands by 1 foot - how high off ground?

[jwpaine]

What should I use from my toolbox to solve this?

A railroad track runs straight and level for exactly one mile. Assume that its two ends remain fixed, but the length of track expands by one additional foot in the summer heat, buckling up into the shape of a circular arc. How far is the middle of the arc off the ground?

I'm asking for pretty much a one word hint. As height increases, arc length gets longer. When arc length = 5281 feet, height = ?

Not sure what to use to solve for this.

 

[galactus]

Hey JW:

Kind of rough to give a one word hint.

The solution to this is rather 'anti-intuitive'. What I mean is, it is larger than you would think.

Let's use three formulae. The arc length formula, the formula for a long chord, and the middle ordinate formula.

The long chord is \(\displaystyle \L\\5280=2Rsin(\frac{x}{2})\)........[1]

The arc length formula is \(\displaystyle \L\\5281=Rx\).........[2]


Solve [2] for x and sub into [1], or any variation thereof:

\(\displaystyle \L\\5280=2Rsin(\frac{5281}{2R})\)

When we solve for R we get 78335.08

Therefore, \(\displaystyle \L\\x=0.0674155..... \;\ rad\) or \(\displaystyle \L\\3.863 \;\ degrees\).

Now, use the formula for a middle ordinate (the distance from the chord to the curve, the height it rises off the ground).

\(\displaystyle \L\\78335.08(1-cos(3.863/2))=\)44.5 feet

See?. It's rather large, huh?. Wouldn't think so.

 

[jwpaine]

Hey JW:

Kind of rough to give a one word hint.

The solution to this is rather 'anti-intuitive'. What I mean is, it is larger than you would think.

Let's use three formulae. The arc length formula, the formula for a long chord, and the middle ordinate formula.

The long chord is \(\displaystyle \L\\5280=2Rsin(\frac{x}{2})\)........[1]

The arc length formula is \(\displaystyle \L\\5281=Rx\).........[2]


Solve [2] for x and sub into [1], or any variation thereof:

\(\displaystyle \L\\5280=2Rsin(\frac{5281}{2R})\) <----- how do you solve for R here? Do I have to use newton's method? Is there any way to solve for R algebraically without newton's method? My TA said newton's method was the only way

When we solve for R we get 78335.08

Therefore, \(\displaystyle \L\\x=0.0674155..... \;\ rad\) or \(\displaystyle \L\\3.863 \;\ degrees\).

Now, use the formula for a middle ordinate (the distance from the chord to the curve, the height it rises off the ground).

\(\displaystyle \L\\78335.08(1-cos(3.863/2))=\)44.5 feet

See?. It's rather large, huh?. Wouldn't think so.

 

[Deleted member 4993]

For a first estimation assume:

sin(x/2) = x/2 - x^3/48

Then {[2] - [1]}

1 = R/24 * x^3...[3]

[2]/[3]

5281 = 24/x^2

x = 0.067413603 rad...........................there .... no invoking Newton