Hanging it Up

I was thinking again about the bent rail problem from a couple of days ago, and got to wondering what the height would be if it was an inverted catenary arc instead of a circular arc.  I was led to this idea by one of my students who also had suggested that it might NOT be circular....and I had no way to assure him it would be... I have no reason to assume it is a catenary either, but it was a curve I felt I needed to know more about so I used it as a self-teaching exercise.

At that point I realized I didn't know much about the catenary curve in general.  (what I mostly knew is about its history and can be found here).  So I set out to see if I could learn a little more and figure out how to solve the question of how high would the inverted catenary reach.  Below is what I have attempted to do... the answer is close enough to the circular amount of bowing that I wonder if it can be right, so if you are up on this stuff, please comment.

I knew that the catenary curve y= a cosh(x/a) has a minimum at (0,a).  For students who haven't seen "cosh" that's just a hyperbolic cosine function (much like the circular cosine function in many ways, and often pronounced to rhyme with "gosh"... see an explanation of hyperbolic functions here).  It can be replaced with an alternate form that may be more palatable to your eyes:

Of course it is easy to see that the vertex is at (0,a).  What I had never known, and find particularly wonderful about the curve, is that for any catenary, the area under the curve from x=c to x=d  divided by the constant a of the equation is the length of the curve.....A good calculus student can show this is true pretty easily with the exponential form since the derivative of cosh(x) is sinh(x).... and for hyperbolas we have cosh^2 - sinh^2 = 1 instead of the addition. That makes the integral for arc length the same as the integral for area between the curve and the x-axis.

So the length of the catenary from the vertex to any other value of x is given by a* Sinh(x/a)  (OK, you can guess that is the hyperbolic Sin, usually pronounced like "cinch" and it is the same as the hyperbolic cosine except the two exponentials are joined by subtraction rather than addition..)

I worked the problem out upside down because I wasn't smart enough to turn it rightside up... ... I assumed that the rod was hanging under a secant line.  For our hanging (and now very flexible) rod, the arc length is 5281 feet, and the secant (parallel to the x-axis) is 5280 feet.  We can position the end points so that they are symmetric around the y-axis so that our hanging rod will have it's vertex on the y-axis. 

So we know that the vertex is at (0,a) and we know that the length of the cable from the vertex to x=2640 is going to be 2640.5 which must equal a*sinh(x/a).    This seemed to give us enough to write an equation that we could solve for a, and my answer came out close to 82,809... which seems like a really big constant.  Buty the effect of a on the catenary is to adjust both the distance from the origin, but also how fast it turns up.  Here is a graph showing catenary curves for a=1 (red) and a=2(blue). 

So we want a very wide catenary, we need a really big number... maybe 82,809 will work.  So our equation for this curve is y = 82,809 * cosh(x/82809).   Now we can use the equation for the curve to find the height of the curve at the point where x= 2640.  This turns out to be just a smidge over 82,851 or almost exactly 42 feet above the y-intercept of 82,809.  So if we flipped it over and set it on the ground, the one mile wide arch would be only 42 feet high.   Very near to the 45 foot answer we got from the circular arc. 

Recently I wrote about Willima Whewell's role in the creation of the word scientist. I just found that there is also a Whewell equation for the catenary which gives the slope of the tangent of the catenary as a function of the distance from the origin, s. Tan(T)= s/a. That means in the catenary version of our inverted rail, the angle between the curve and the horizontal secant is just over .03 radians.... I think I did that correctly.

As a comparison, the St. Louis arch is 630 feet high, and has a width from side to side of 630 feet.  It is not a true catenary because it is adjusted for weight, called a "weighted" or "flattened" catenary. If we assume the difference is small (I don't know for sure) then the information above should allow you to find the constant "a" for the arch, and then find the length of the arch... Good luck... 

Oops, Not Quite to Specification

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

And of course the tougher challenge would be to calculate the height exactly (or at least some reasonable accuracy)..

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

S
P

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L
E
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 DE² = 2640.25² - 2640² . 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 - x³  /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  H² = {(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 - θ³/6,  we get  r² @ (L + ΔL)³/ 24 ΔL.  We have  H = r (1 - cos θ) @ rθ²/2  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.