Thanks much! An answer I can understand. Also gratified to see that my approximation of 14.5' was not THAT far off (I did say it would be "a little higher" than the actual value).
I think I previously laid out the problem in a similar way, but got stuck on the "trial and improvement" step.
Glad I can close out this issue which has been left unfinished since college. Now, to find a ghost writer for my novel ... :-)
I think there is only way a steel bar can be when bent like that, but I will leave that up to physicians.
What do medical doctors know about steel bars?
An actual steel bar would most likely just distort or compress if you were able to apply sufficient pressure to one end to push it inward by a foot; or bend in the middle so it was on the ground anyway. Unless you did it in zero G.
Anonim, where do I look for your answer?
Aha. I had always assumed it was one of those problems which seems complex on the outside but has a simple answer if you could just see it. Looks like it's just the opposite - seems simple but it ain't.
I last took numerical analysis in 1984; didn't expect to need it for this problem! But even in those days, we had access to some pretty decent computers; didn't have to rely on the Radio Shack calculator.
A friend on the fencing team posed this problem to me, 30 years ago; I've considered it off and on since then. I have an APPROXIMATE answer, but I've never been able to solve it rigorously.
Say you had a perfectly straight steel bar a mile long. You push in on one side by one inch, so it bows in the middle. Would the resulting gap at the center of the bar be high enough to ...
- slide a penny under it?
- roll a baseball under it?
- drive a truck under it?
I answered "all of the above," and my friend said I was correct. Then came the sticking point -- PROVE IT!
Since then, I've tried a variety of approaches on my own; made great circles in which the bar was an arc and the ground was a chord; assume it's a parabola and try to solve for H at the midpoint (but that leads to a complex expression using hyperbolic sin to get the length of the parabola). The best I can do is to break the bar into two straight line segments, and solve for the height of the isosceles triangle - hypotenuse 2640 feet, one side 2639.96 feet:
2640^2 - 2639.96^2 = H^2
211.2 ~= H^2
H ~= 14.5
That's a little higher than the actual answer, since the peak of the triangle will be higher than of an arc of the same length. But not much.
How would you get a more precise answer?
50 year old Marathon runner whose heart goes out to the runners, families, and friends at the Boston Marathon.