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...