One inch caused that???

Frank Stewart writes a Bridge Column for the Philadelphia Inquirer. In the 17 January 2001 edition, he included a Mathematics problem. The answer to the problem seems incredible to me, and I wonder if Frank misinterpreted his source or a typo crept into the paper. The problem is as follows

Quote:

---

Imagine a 3000 mile long strip of metal laying flat. With the ends anchored, suppose the length expanded one inch. how high would the center of the strip rise due to the increase in length?

---

He claimed that it would rise 800 feet!!!

I did some calculations and came up with a very different answer, more in line with my intuition. It is, of course possible, that my analysis and/or computations are incorrect.

Have any of you encountered this problem? Can anybody come up with an answer of their own? I do not want to influence your thoughts by posting my analysis or numerical result.

I would do this by approximation to check a 'ball park figure'. Assume that the metal strip rises in a triangular fashion. This gives two right angled triangles. The base of each is 1500 miles and the hypotenuse is 1500 miles and half an inch. What is the height? Intuitively a lot less than 800 feet.

Now, if the base is x, the height is y and the hypotenuse is z then x^2 + y^2 = z^2 => y = (z^2 - x^2)^0.5

In inches:
x = 95,040,000. z = 95,040,000.5
x^2 = 9,032,601,600,000,000
z^2 = 9,032,601,695,040,000.25

=> y^2 = 95,040,000.25
=> y = ~9,748 = 812 feet. This is the upper limit. This would be reduced markedly by the metal forming a curve. This is anti-intuitive, but it doesn't seem too far away...

Now let me know what you got.#

Cheers,

P.

i got 124m ~406 feet assuming it would rise in a arc of a circle with the radius 2 614 877 064 m

kedaman,

What formula did you have to calculate the radius? I looked at it graphically with a chord and an arc but I didn't give it too much thought... Just interested as to your approach.

P.

Actually, i used the equation solver on my Ti83 to get the angle of the circle, from the formula in bold:

b=arc
a=angle
s=secant
r=radius
m=r-secgment height

b=ra, sin(a/2)=2/r^2 -> b=as/(2sin(a/2))

m=sqr(r^2-s^2/4)

a=6.155e-4 radians ~127 seconds

r=2 614 877 064 m ~ 4 times sun radius

r-m=124m ~ 406 feet

hmmm. OK. Not a lot wiser, but never mind. Interested to see what Guv has to say...

P.

uhm what do you mean not a lot wiser?

It's seems reasonable you have only 1 inch for a 1/4 of earth radius to bend only 124 m. That's a ratio of 13000:1. OKOK, I guess you could never know what would be, if you have these dimensions...

My original answer was about 2 inches. I now believe that 703 feet is a correct answer. I never make mistakes, but when I do, they are not trivial errors.

It never occurred to me to think of a 3000 mile long strip of metal lying flat, like in a straight line from plane geometry. I assumed it was lying on the surface of a sphere 4000 miles in diameter (an idealized version of the earth).

The above assumption leads to a height of about 2 inches for the slightly longer strip. Since this agreed with my intuition, I assumed that the Bridge Columnist made an error. I even assumed that the actual problem would have a similar solution. It shows how good my intuition is. Oddly enough, the equation I had to solve is the same equation used below (with different constants) for the problem as stated by Frank Stewart. After solving it, I then calculated the distance between two circular arcs, rather than the distance between an arc and a chord.

PaulW: It was a cute idea to get an upper bound by assuming a right triangle. I think this might be the solution the Bridge Columnist had in mind. Using your method, I got the same 812 feet (approximately) that you did. Your idea made me realize that the strip being a straight line leads to a much bigger height than assuming it is an arc of a circle. Perhaps you got the idea because you are a member of the Flat Earth society.

Kedaman: I do not understand your equations. This is probably because I am not looking at whatever diagram you used to arrive at them. Too bad it is difficult to post sketches and diagrams for this sort of thing. If you used a hand calculator, our results might differ due to differences in the precision of the computations. My MathCad7 software can work with the precision of a VB Double. Maybe you can understand my equations without a diagram and tell me that they are equivalent to yours. You are obviously working with an angle which is twice the angle I am using, and I can picture using your angle to arrive at a solution.

Everybody: After reading PaulW's Post which got me on track to solving the problem as stated, I used the following approach to the problem.

1500.000007891414 = 1500 miles plus one half inch.

HalfArc equation: 1500.000007891414 = Radius * Angle (Angle in Radians)
HalfChord equation: 1500 = Radius * Sin(Angle)

Radius = 1500.000007891414 / Angle (Rearranged HalfArc Equation)

After substituting for Radius in the HalfChord equation, and a little manipulation, The above leads to the following

F(Angle) = 1500.000007891414 * Sin(Angle) - 1500 * Angle

When F(Angle) = 0, the conditions are satisfied. I got the derivative and used the Newton method to find a solution. BTW: The Newton method is a damn handy gadget to keep in mind.

I came up with the following.

Angle = 0.000177667261147 Radians (0.010179584221372 Degrees; About 37 Seconds)

Radius = 8,442,748.530075723 Miles

Final Answer: Height = 5280 * Radius * (1 - Cos(Angle)) = 703.5623569833184 Feet.

I used my MathCad7 software for the above, and feel fairly confident in the results. Considering that the PaulW value (812) is an upper bound, the above (703) looks about right.

It really is startling and counter-intuitive. I hope there are no typo's in the above.

Chord, i suppose this is "Secant" i'm probably the one using the wrong word...

Quote:

---

1500 = Radius * Sin(Angle)

---

This doesn't seem correct, this would be on the other hand:
1500/2 = Radius * Sin(Angle/2)
since we have this diagram:

Code:

---

|a\ngle/2
|  \
|  \radius
|    \
|    \
| _____\
 chord/2

---

otherways i think we are doing about the same calculations, except that i don't know about the Newton method.May I ask what is it good for?

Your radius is 5 times mine, i don't know how you got that, probably because the angle got so small..

My Ti83 has about double precision too, 14 decimal and float values from e-99 to e+99, but i'm not sure if the equation solver forced that exact results.

The newton method can be used with almost any function for which you can compute the derivative. The basic approach is the following.

You want to find a value for which some function equals zero. You make a guess and use the following.

Next = Last - Function(Last) / Derivative(Last)

If Last is a guess, then Next is a better value. The following is a very simple example for finding the cube root of 1000 (which we know is 10).

Function = X^3 - 1000
Derivative = 3 * X^2
First bad guess is 20

20.000000 = First guess.
14.166667 = 20.000000 - (20.000000^3 - 1000) / 3 * 20.000000^2
11.105344 = 14.166667 - (14.166667^3 - 1000) / 3 * 14.166667^2
10.106368 = 11.105344 - (11.105344^3 - 1000) / 3 * 11.105344^2
10.001116 = 10.106368 - (10.106368^3 - 1000) / 3 * 10.106368^2
10.000000 = 10.001116 - (10.001116^3 - 1000) / 3 * 10.001116^2

The above example shows numbers rounded to 6 decimal digits.

The precision tends to double for each iteration. If you start with a really bad guess, it takes a few iterations to get close and then you zoom to the correct value.

You can understand the method geometrically if you visualize the curve (Y = X^3 - 1000) in the XY-Plane.

The tangent to that curve at 20.000000 intersects the X-Axis at 14.166667
The tangent to that curve at 14.166667 intersects the X-Axis at 11.105344
The tangent to that curve at 11.105344 intersects the X-Axis at 10.106368
The tangent to that curve at 10.106368 intersects the X-Axis at 10.001116
The tangent to that curve at 10.001116 intersects the X-Axis at 10.000000

There are funky functions which cause the method to jump around and get nowhere. This only happens when there are multiple solutions to the equation you are working with, and those solutions occur at "unlucky "places." You can visualize the problem by imagining the geometric description applied to a polynomial with many roots, and an X-Value which results in a tangent parallel to the X-Axis and above the X-Axis. A guess on one side of the relative minimum gets you to the other side of it, and vice versa.

I apologize if there are any typo's in the above.

Thanks Guv, that could be handy :)

It's also called the Newton-Rahpson method, or at least that's what I know it as.