Railroad Track Expansion
Date: 04/07/2003 at 00:41:34 From: Jerry Looney Subject: Arcs and Chords I read the teaser in the "Ask Marilyn" column that presented the problem of expansion of a railroad track. A continuous straight railroad track of one mile is permanently tied down at both ends so that neither end can move from its tie-down point. As the day heats up, the coefficient of expansion of steel causes the rail to expand a total of one foot over the entire length of track so that the length is now 5281 instead of 5280 feet. Assuming that the track expands in an upward direction, what maximum vertical distance from the horizontal will the track rise at the highest point? The choices given for the answer were one inch, one foot, or greater than 50 feet. The answer came back this week as over 50 feet and I have not been able to solve this puzzler. I used the formulas for segments of a circle and came up with .52 feet. Perhaps I need to brush up on math that I used 50 years ago or maybe by some miracle their answer was wrong. My daughter and son-in- law with masters degrees in engineering both guessed one inch also.
Date: 04/07/2003 at 10:21:22 From: Doctor Rick Subject: Re: Arcs and Chords Hi, Jerry. Did you use the formulas from the Dr. Math FAQ? Segments of Circles http://mathforum.org/dr.math/faq/faq.circle.segment.html We know the chord length (c) and the arc length (s) so we need case 1, which is one of the more difficult - it requires numeric methods. When I follow that method using a spreadsheet, I get c = 5280 feet s = 5281 feet x = 0.033707759 theta = 2x = 0.067415518 radians r = s/theta = 78335.0805 feet d = r*cos(x) = 78290.58205 feet h = r - d = 44.49845502 feet That's less than 50 feet, so on the assumption that the expanded track forms an arc of a circle, Marilyn's answer isn't quite right, but it's pretty close. We can check Marilyn's answer more easily than we can come up with our own. If the chord is 5280 feet and the height at the center is 50 feet, we can use case 8: r = (c^2+4h^2)/(8h) = 69721 feet theta = 2 arcsin(c/[2r]) = 0.07574852 radians s = r theta = 5281.2625 feet That's a bit more expansion than the problem indicated, so the actual height (assuming a circular arc) would be a bit less, not more. Now I'll comment a bit on how we can make sense of this counterintuitive result without relying on such complicated calculations. Let's make the problem simpler. Suppose that, instead of forming an arc of a circle (which is probably not accurate anyway since it would probably form a catenary arch), the rail kinks at the center, forming two straight segments. Then we have two back-to-back right triangles. Each hypotenuse is 5281/2 = 2640.5 feet, and each horizontal side is 5280/2 = 2640 feet. What is the third side of each right triangle - the height at the center? Use the Pythagorean theorem: b^2 = c^2 - a^2 = (2640.5)^2 - (2640)^2 = 2640.25 b = sqrt(2640.25) = 51.38 feet This is understandably on the high side. An arc with this height will be longer than the sum of the straight segments, since a straight line is the shortest distance between two points. But we have a reasonable order-of-magnitude approximation. In fact, it may be that Marilyn's calculation was on this assumption - a kink in the track, rather than a circular arc - because this answer is indeed a bit more than 50 feet, as she said. To deal with your intuition, consider some right triangles with one leg much shorter than the other. You'll see that you need to increase the short leg quite a bit to make the hypotenuse significantly greater than the longer leg. Does the answer make sense yet? Can you see what you and your family engineers did wrong? - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 04/07/2003 at 11:52:14 From: Jerry Looney Subject: Thank you (Arcs and Chords) You guys are great with explanations and teaching. I am going to forward this to my daughter (MS from Purdue in engineering) and son- in-law (college educated in the UK) and inform them that they are just as rusty as their old dad/father-in-law with their geometry and to make sure they brush up on geometry before they continue their work on designing Ford automobiles in Detroit ;-)) Thanks a bunch and am glad to know about this website, Jerry Looney
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.