If you enjoy sharing problems with your math classes, here is one that will challenge the thinking skills of your upper level students. Suppose a perfectly flat piece of ground is prepared that is one mile long. A thin rail of steel is ordered to go across the flat surface. Unfortunatly, the bid went to the lowest bidder, and they were not very good with a ruler, so the rail was 5281 feet long instead of 5280. The engineers decided to squeeze it into place and let the middle bow up as much as necessary. And the question is.... How much did it bow up?

The intuitive approach is to estimate whether the hight above the ground in the center of the rail would be

a) just enough to slide a piece of paper under
b) just enough to for a small dog to walk under

c) just enough for the average man to walk under
d) just enough for the average car to drive under

e) just enoug for a London double-decker bus to drive under

Stop now if you don't want to see the solution... or read on for the solution and a related question....

S

P

I

LE

R

If we assume that the bowed rail makes a circular arc, as shown, then we can apply some of the geometry we have learned to get a pretty good (I hope) approximation... but first we might try to find a back of the envelope aproximation that would answer the multiple choice portion:

If we tried to create the right triangle shown at left we can hope that with some reasonable estimates we could come up with a (very) rough value for the bowed distance DE. DC is 1/2 mile or 2640 feet, and the Arc from C to E is 1/2 a foot longer, or 2640.5 feet. So what should we say about CE, the hypotenuse of the right triangle. Can we suggest 2640.25 as a compromise length. If so, then by the Pythagorean theorem we can obtain DE

^{2}= 2640.25^{2}- 2640^{2 }. From this we can estimate that the height of the bend is about 36 feet... We could get slightly different estimates for the height for different estimates of CE, but even if we assume it were only 2640.1the height to the bend is over 20 feet.
The biggest problem in a more exact solution is that we come up with a couple of messy equations in two unknowns (or at least I do)... I will call the angle CAD simply angle A for typing ease, and we have from two common formulas from analytic geometry that the arc length is equal to r times the central angle. For our problem, the arc CE will equal A*r. We can also use the right triangle ADC to show that the Sin (A) = 2640/r... or r * Sin(A) = 2640...

Now that is pretty messy to try to graph on a calculator, and it turns out that Wolfram alpha folded on the problem....

At this point they have a perfect place to apply the series approximations to Sin(x) that you made them learn... ie that Sin(x) = x - x

^{3}/6 ... and if the angle A is even less than 1 radian, we can assume our error is less than 1/120 radians... not a terrible estimate.
That allows us to write an expression for Sin(A)/ A that will simplify into something tractable.

From this we can conclued that A must be about .0337... and from that we can conclude that the radius must be about 2640.5/.0337 or about 78,350 feet . Now we can subtract the distance AD which is r Cos(A) or 78,305 to get a (hopefully) close approximation of the height of the bowed rail of 45 feet... room for that double-decker bus for sure...

And the follow up question... What does that make the actual length of that hypotenuse we were estimating earlier... and is there a pretty pattern in there somewhere about the length of a chord... no time to think...kids coming in the door..."Run away.. ...run away!"

-------------addendum---------------

Jeffo has pointed out that if you used the equation Sin(A)/A above instead of the system of equations then Wolfram alpha will solve the value of a, and gets the same value, but with much greater accuracy....

A ~~ ±0.03370775880988154948897645696380541681505...

Thanks, Jeffo.

Addendum two::: The history of this problem has been well documented by David Singmaster. This is from his notes

A railway rail of length L and ends fixed expands to length L + ΔL. Assuming the rail makes two hypotenuses, the middle rises by a height, H, satisfying H^{2}= {(L+ΔL)/2}^{2}‑ (L/2)^{2}, hence H @ Ö(LΔL/2).However, one might assume the rail buckled into an arc of a circle of radius r. If we let the angle of the arc be 2θ, then we have to solve rθ = (L + ΔL)/2; r sin θ = L/2. Taking sin θ @ θ - θ^{3}/6, we get r^{2}@ (L + ΔL)^{3}/ 24 ΔL. We have H = r (1 - cos θ) @ rθ^{2}/2 and combining this with earlier equations leads to H @ Ö{3(L+ΔL)ΔL/8} which is about Ö3 / 2 = .866... as big as the estimate in the linear case.The Home Book of Quizzes, Games and Jokes. Op. cit. in 4.B.1, 1941. P. 149, prob. 12. L = 1 mile, ΔL = 1 ft or 2 ft -- text is not clear. "Answer: More than 54 ft." However, in the linear case, ΔL = 1 ft gives H = 51.38 ft and ΔL = 2 ft gives H = 72.67 ft, while the exact answers in the circular case are 44.50 ft and 62.95 ft.Sullivan. Unusual. 1943. Prob. 15: Workin' on the railroad. L = 1 mile, ΔL = 2 ft. Answer: about 73 ft.Robert Ripley. Mammoth Believe It or Not. Stanley Paul, London, 1956. If a railroad rail a mile long is raised 200 feet in the centre, how much closer would it bring the two ends? I.e. L = 1 mile, H = 200 ft. Answer is: "less than 6 inches". I am unable to figure out what Ripley intended.Jonathan Always. More Puzzles to Puzzle You. Tandem, London, 1967. Gives the same question as Ripley with answer "approximately 15 feet". The exact answer is 15.1733.. feet or 15 feet 2.08 inches.David Singmaster, submitter. Gleaning: Diverging lines. MG 69 (No. 448) (Jun 1985) 126. Quotes from Ripley and Always.David Singmaster. Off the rails. The Weekend Telegraph (18 Feb 1989) xxiii & (25 Feb 1989) xxiii. Gives the Ripley and Always results and asks which is correct and whether the wrong one can be corrected -- cf Ripley above.Phiip Cheung. Bowed rail problem. M500 161 (?? 1998) 9. ??NYS. Paul Terry, Martin S. Evans, Peter Fletcher, solvers and commentators. M500 163 (Aug 1998) 10-11. L = 1 mile, ΔL = 1 ft. Terry treats the bowed rail as circular and gets H = 44.49845 ft. Evans takes L = 1 nautical mile of 6000 ft and gets almost exactly H = 50 ft. Fletcher says it took 15 people to lift a 60ft length of rail, so if someone lifted the 1 mile rail to insert the extra foot, it would need about 1320 people to do the lifting.

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## 5 comments:

I got the same answer, but with a different approach. I put A at the origin and D on the positive x-axis. Then the equation (sin x)/x = 2640/2640.5 falls out with simple trig. WA can handle that (with FindRoot) quickly to get x = 0.0337.

Jeffo,

I actually only tried them with the system, never even tried to use the simpler single variable equation.. thanks, will make a note on the blog...

Yes, now that I'm looking at it, your system was equivalent to mine. . .

I think you need to take the compression of the steel into account. A mile of steel rail would be pretty heavy and at that small angle (if it could be held perfectly in a line above the ground) it would probably compress itself until almost flat...

Thanks for this interesting geometry problem which shows how circular lengths can only be compared approximately to rectilinear lengths, except if the angle has a sine which we know precisely (60, 45, 30, ... deg). If we had arc = chord * pi/2 (rad), or 2*pi/(3*sqrt(3)), ... the height can be determined to the precision of which we know pi or square roots. I don't know about the compression of steel, but as it is a thin rail of steel, it is probable that it will bend. Maybe a physically more probable solution would be that the rail bends parabolically. Parabolic segments and heights can be calculated more precisely than circular segments and heights, so this would yield a more precise height.

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