OK, that gives me 51ft which makes a bit more sense (not much, but a bit). How would you calculate the actual height?
How much the mile of railroad track with the extra foot “bows up” by depends upon the shape it assumes. If, to make things simple, we assume that it forms a triangle with a sharp bend in the center (I know it’s not real, but it’s easier to solve), then it’s just the Pythagorean theorem, and the maximum height is one half the square root of ( two times the original length times the added length). Since the original length is 5280 feet and the added length is one foot this gives 51.4 feet, which is a little surprising, but not as bad as 1056 (which is almost a quarter of a mile!)Of course you’d get a smooth curve, but It’s been awhile since I did that kind of calculation. 51.4 feet will, in any case, be greater than the answer you’d get with any gentle curve, and ought to be correct to the order of magnitude. (From my experience with bending beams you won’t get an arc of a circle, or a sine curve. I think the answer is a cubic – but the differences between these aren’t all that large. I’ve neglected a term in delta squared, again, but it’s one part in 5280, so it’s not significant.)
“Tell Zeno I’m willing to meet him halfway.”
Holy shit! Now i know why i got a C in Algebra. 
The intended assumption for “bows up” was to be a circular arc. The triangle assumption illustrates the point. I brought this thing up once, and somebody was going to try it adding 1 mm to 15 meters of fishing line because they didn’t believe it. Of course, that’s impractical - when you straighten out the fishing line, you’ll have to tension it to some degree, and it will have a lot more stretch than 1 mm in it.
My detractors are correct. I am a Nerd.
For absolutely no good reason I have worked out how much the rail bows out if you assume that the shape is an arc of a circle. It works out to be about 44.5 feet. This is less than the 51.4 feet I gor with the triangular approximation, as I said it would be. I refuse to try and write the result out until you people get an equation editor.
I’m going back to the threads where you talk about movies, or swabbing hot oil over the other SDMBers
Here’s my answer to the OP. For once i did it all in one go and I believe it to be correct. maybe someone can double check for me.
To clarify: we are looking for the separation from the surface of the globe of a tangent to it at a point at a distance L from the point of tangency.
Call R the radius of the Earth.
Call L the great circle distance between the point in question and the point of tangency.
Call Alpha the angle subtended by L from the center of the sphere
Call H the excess distance we are looking for
This is a very easy geometric problem as you have a rectangular triangle where the hypothenuse is R + H , another side R and another side R/cos(alpha)
H is therefore equal to hypothenuse minus R
H = R * ( (1/cos(alpha)) -1 )
to have alpha expressed in radians take L and divide by the radius of the sphere, so you have:
H = R * ( (1/cos(L/R)) -1 )
This formula, if I have not made any mistakes, should be exact. Using 20887000 ft as the radius of the Earth and a spreadsheet to calculate I got:
L' H' H"
100 0.0002 0.0029
250 0.0015 0.0180
500 0.0060 0.0718
1000 0.0239 0.2873
2500 0.1496 1.7954
5000 0.5985 7.1815
10000 2.3938 28.7260
25000 14.9615 179.5376
50000 59.8460 718.1518
100000 239.3856 2872.6276
250000 1496.2352 17954.8229
500000 5986.0130 71832.1557
1000000 23961.2190 287534.6285
I suppose the numbers will not line up but I don’t know how to make them do it.
>> Assume you have a mile of straight railroad track tied down at each end. Somebody welds an extra foot of track into it, causing it to bow up in the middle. How far out does it bow?
I am assuming the track takes the shape of a circular arc (if we assume the triangular shape the problem is trivial). As I do not remember the number of feet in a mile I am assuming 5280 (I am metric). The reasoning should be good, just correct the numbers.
Assuming the circular arc then we can easily summarize the problem as follows:
We have a segment of a circle with a length of 5281’ and the chord is 5280’. The first thing we do is find out the radius of the circle. We call (2alpha) the angle subtended by the arc (alpha in radians), R the radius of the circle, (2L) the length of the cord (5280’ here), (2*A) the length of the arc (5281’ here) and D the distance we are looking for.
L = R sin(alpha)
A = alpha * R
D = R * (1 - cos(alpha) )
From that we get A / L = alpha / sin(alpha)
We have A / L = 5281 / 5280 but I do not know how to resolve alpha from this equation except by succesive approximations using a spreadsheet. Doing that I get:
0.03370774 radians (= 1.9313 degrees) which, in turn gives
R = 78335’ and D = 44.5’
Let us double check by doing it backwards… yup, it works so the answer is 44.5 inches of mercury 
PS: stop posting these problems because I am compulsively forced to resolve them.
CalMeacham sez:
(You can find the story in Martin Gardner’s “Fads and Fallacies in the Name of Science.”
Alright! Someone else quotes one of my favorite books! And I was thinking I was the only Doper who’d read it.
Good job, Sailor, 5,280 is, in fact, correct. If you’re going to pick and choose a shape to make the problem simpler, though, why not say that it bows out in a rectangular shape? Then, the answer is six inches exactly
Also, anyone want to try to figure out the actual shape of such a bow would be? Minimize total energy; assume that the steel is incompressible, but flexible, and that the potential energy of the flex is proportional to the integral of the curvature along the length of the beam. Don’t forget gravitational potential energy, and judging from the answer to the OP, we shouldn’t neglect the curvature of the Earth on this scale, either. Have fun! 
>> If you’re going to pick and choose a shape to make the problem simpler, though, why not say that it bows out in a rectangular shape?
Chronos, get real. Neither rectangular or triangular shapes come even close to reality. A circular arc, while not the true shape, comes much closer to approximating it.
I am assuming the ends are free. In other words, imagine the rail is 5281’ lying freely on a flat surface and we attach a steel cable to both ends and shorten the distance by 1’.
If this is the case, common sense tells us that:
(1) If the arc were a half circle, then the arc would indeed be a segment of a perfect circle.
(2) If the arc is less than 180 (as is this case where it would be under 2 degrees) then the curvature is greatest at the center and diminishes gradually towards the ends. This means the true answer would be somewhat greater than what I calculated (44.5") It is very possible that the shape would be a parabola or hiperbola but I am not going to try to figure it out.
If you define the problem differently, like the ends of the rail are inserted in two walls that an earthquake brings closer by 1’, then the problem is very different as the rail would take a sort of omega shape, or sine function shape, as both ends have to keep pointing in the original direction.
You all realize that these questions are so basic any student of mechanics can resolve them easily. They are representative examples of questions asked in tests.
In any case, I’ve done my part and someone else can take over. I’m done.
The shape you’re looking for is a catenary. This is the shape a piece of string will assume if it’s dangling freely from its two ends.
It’s also seen as the kind of arch you would use to support a bridge. And the St. Louis arch.
You may also have seen it in school or at a science museum where blocks are stacked into an arch that supports itself without glue.
The equation is y = a cosh(x/a).
vd: I’ve heard that before, that a bent flexible rod will assume the shape of a catenary, but it doesn’t work. You can tell this by considering the extreme case: If you bring the ends of a chain together, you find that the catenary degenerates into a line. On the other hand, if you take a flexible rod and bring the ends together, you get a sort of teardrop shape. As an aside, bridge arches are typically parabolas, not catenaries, depending on the distribution of the weight they’re supporting.
sailor: I was just being facetious, of course, about the rectangle. I could solve the problem more realistically, but I don’t want to waste any of my own time on it :). You’ll note that I phrased the problem in such a way as to make it a bit more difficult than the standard textbook cases, too. I wouldn’t exactly call calculus of variations “basic”, either… My definition of “basic” would be at a college freshman level.
VD. I believe you are mistaken about the catenary. The catenary assumes no rigidity (among other things).
Chronos, I meant “basic” for a mechanical engineer and I further meant that I don’t have a clue why I would be wasting my time on this which is pretty difficult and pointless and yet a mechanical engineer could do it with his eyes closed.
Some more useless trivia about curves: I once designed a bed headboard cut in the shape of an ellipse. When I drew the curve on the board for the guy who was going to build it for me, he was quite surprised and told me he loved it and was going to build one like that for his daughter. I used the old method of a length of string that goes to two nails at the focii.
Another curve I like (but have not used) is Piet Hein’s Super Ellipse
another Super Ellipse site
The Superellipse makes a very pleasing curve for furniture (tables), gardens etc.
Other Curves
Sailor:
I’ve been a big fan of superellipses since I read about them in Martin Gardners MathematicalGames column ages ago.
Once you know about superellipses you can extend the field to other curves, including the superLorentzian and the superSquareWell (which is simply x to the Nth power). I also derived a formula for a triangular form of arbitrary shape. As you increased the power of the curve, it more and more closely approached a traingle.
sorry about that link
Superellipse
At the Curves site you can select a curve and then select a java applet that will plot it with any parameters you give it.
Your racking YOUR brains out and calculating FOR HIM on the back of an envelope and HE needs therapy?!?! L
>> By my back of the envelope calculations the mythical board would be a little over five miles long
Padeye, there’s something wrong with that envelope. As I showed 5000 feet (undeer a mile) give you more than 7"