[Prof. Templeton]

A variation of an earlier one...

The employees of Saudi Aramco Oil company were charged with installing a steel pipe 5km across a flat desert from the fields to the refinery. The temperature in the desert reached a scorching 105 degrees F during the day, so it was decided to install the pipe at night when the temperature was a cool 45 degrees F. The 5km of steel was laid on top of the flat desert sand all in one night and both ends securely fastened at their respective terminations, but nowhere else. The next day the employees found that the steel had expanded in the 105 degree F heat and bulged straight upward in the center of the 5km line forming a circular arc.

Steel expands .00000656 inches per inch per degree F.

Disregard pipe diameter for this problem and treat as a line.

How many meters above the ground was the center of the pipe?

This is only a puzzle and not real life.

[Guest]

Just a guess, but:

The pipe is 196,850 inches long. The pipe would expand 1.29 inches for every degree in temperature change. Since it rose 60 degress, the pipe expanded 77.48 inches, or just shy of six and a half feet. Or did I do my math wrong..

[Prof. Templeton]

Just a guess, but:

The pipe is 196,850 inches long. The pipe would expand 1.29 inches for every degree in temperature change. Since it rose 60 degress, the pipe expanded 77.48 inches, or just shy of six and a half feet. Or did I do my math wrong..

The 5km pipe did get 77.48 inches longer. But that's not the question.

[Guest]

Just a guess, but:

The pipe is 196,850 inches long. The pipe would expand 1.29 inches for every degree in temperature change. Since it rose 60 degress, the pipe expanded 77.48 inches, or just shy of six and a half feet. Or did I do my math wrong..

not all of the increase in height goes towards displacing the center of the pipe. Because the pipe is an arc after expanding, your guess is an over estimation. However, I have not had time to work through the arc length equation to figure out the actual height.

[Guest]

not all of the increase in height goes towards displacing the center of the pipe. Because the pipe is an arc after expanding, your guess is an over estimation. However, I have not had time to work through the arc length equation to figure out the actual height.

Well. Since the question asked for meters, it'd be 1.97 meters off the ground at the center. Again, assuming that I did my math right overall and I'm not missing something.

[Guest]

So I tried to go through the math but I tend to fat finger things a lot but i got

2391.69 inches?

Edited March 11, 2009 by Orykle

[Guest]

OK, so the trick is to find the diameter of the loop that gets created from the expansion. To do that we find the circumference of the loop, which is the same as the total expansion length:

0.00000656 (displacement per inch per degree F) x 60 (degrees F) x (196850 inches pipe lenght) = 77.48 inches

77.48 inches = 1.97 m (since question asks for meters above ground)

Now to find diameter we do the following:

diameter = circumference / pi

1.97 / 3.14 = 0.62 m

So, not factoring in the pipe diameter the center of the pipe is 0.62 m above ground.

[Guest]

OK, so the trick is to find the diameter of the loop that gets created from the expansion. To do that we find the circumference of the loop, which is the same as the total expansion length:

0.00000656 (displacement per inch per degree F) x 60 (degrees F) x (196850 inches pipe lenght) = 77.48 inches

77.48 inches = 1.97 m (since question asks for meters above ground)

Now to find diameter we do the following:

diameter = circumference / pi

1.97 / 3.14 = 0.62 m

So, not factoring in the pipe diameter the center of the pipe is 0.62 m above ground.

I do not see how this would be the answer. The length after expansion is an arc length, not circumferance. That is, the arc length is a fraction of 2pi*r.

[Guest]

If it forms a perfectly circular bulge I believe it would be 0.626 meters above ground (the radius). I got that by taking the 77.48 inches of expansion, converting it to 1.968 meters which should be 1/2 of the circumference of the circle. So I divided that by pi for the radius which is 0.626meters

[Guest]

So I tried to go through the math but I tend to fat finger things a lot but i got

2391.69 inches?

Agreed, i got a similar answer probably due to rounding. You have to use trig or calculus to solve this problem and not just simple circle equations. Haven't used those math tools in a while

useful equations if you want to solve it for yourself:

http://mathworld.wolfram.com/CircularSegment.html

2392.3 inches or 60.76 meters high

[Guest]

I do not see how this would be the answer. The length after expansion is an arc length, not circumferance. That is, the arc length is a fraction of 2pi*r.

The problem says it creates a circular arc. To me that means the pipe meets at an O (circle) in the middle, and displacement from the ground is the diameter of the circle.

If you are after the arc length (or heigth of the arc) then the question is impossible since we would need to know the width of the arc to solve it. I.e. if the width of the arc is 5 meters then the height will be X, but if the width of the arc is 50 meters than the height will be Y.

But if it truly is a circle then I believe my answer is correct.

[CaptainEd]

Agreed, i got a similar answer probably due to rounding. You have to use trig or calculus to solve this problem and not just simple circle equations. Haven't used those math tools in a while

useful equations if you want to solve it for yourself:

http://mathworld.wolfram.com/CircularSegment.html

2392.3 inches or 60.76 meters high

Apparently Furikan did some additional work beyond what was shown in the site he alludes to--I didn't instantly see a relationship between the arclength and the chord, so I didn't see how he would know the presumed radius of the circle. But since I didn't do the work and he did, I can believe it.

Assume the arc is really just a tent with a pole in the center, whose height we must calculate. The length along the ground is 2500m (half the length of the pipeline), the length of the cloth that is shading the ground is 2500.985m (half of 5000+1.97m). Using Pythagorean theorem, the height is 70.1m. In actuality, the cloth is shaped like an arc, as Furikan points out, so the height will be less. Quite possibly the exact number he gave.

Nice counterintuitive puzzle, Dr. Templeton!

Edited March 11, 2009 by CaptainEd

[Guest]

Agreed, i got a similar answer probably due to rounding. You have to use trig or calculus to solve this problem and not just simple circle equations. Haven't used those math tools in a while

useful equations if you want to solve it for yourself:

http://mathworld.wolfram.com/CircularSegment.html

2392.3 inches or 60.76 meters high

I agree with you. I got just shy of 60m, but I did some lazy rounding.

[Guest]

The problem says it creates a circular arc. To me that means the pipe meets at an O (circle) in the middle, and displacement from the ground is the diameter of the circle.

If you are after the arc length (or heigth of the arc) then the question is impossible since we would need to know the width of the arc to solve it. I.e. if the width of the arc is 5 meters then the height will be X, but if the width of the arc is 50 meters than the height will be Y.

But if it truly is a circle then I believe my answer is correct.

I believe that the arc extends from one end of the pipe to the other.

[Guest]

The solution I think.

Diagram taken from my previous post since my drawing skills are terrible.

Where a is the original length of the pipe.

s is the length of the pipe is the heat.

Both of those are known as stated in previous posts.

Trig will tell you that the length of the known chord, a, is given by the equation: a=2Rsin(theta/2)

You must also use the formula for arc length, s=R*theta, and substitute for R giving: a=2(s/theta)*sin(theta/2)

We now have an equation where the only unknown is theta! Make sure you solve for it in RADIANS!

With theta now we can find R with the arc length formula, R=s/theta <<still in radians.

With R and the the chord length, a, 2 sides of the right triangle are known so pythagorus can tell us the length of the remaining side, r.

r=sqrt(R^2-(a/2)^2)

The relationship between R, r, and h is given by the equation h=R-r

[Guest]

Here is an explanation of how I understood the OP.[spoiler='explanation for answer

'] the pipe at night before expansion is a cord in a circle, the pipe after expansion is the arc enclosed by the end points of the cord. the height asked for is the distance of a line at the center of the cord, perpendicular to that cord, intersecting the arc.

[CaptainEd]

The solution I think.

Diagram taken from my previous post since my drawing skills are terrible.

Where a is the original length of the pipe.

s is the length of the pipe is the heat.

Both of those are known as stated in previous posts.

Trig will tell you that the length of the known chord, a, is given by the equation: a=2Rsin(theta/2)

You must also use the formula for arc length, s=R*theta, and substitute for R giving: a=2(s/theta)*sin(theta/2)

We now have an equation where the only unknown is theta! Make sure you solve for it in RADIANS!

With theta now we can find R with the arc length formula, R=s/theta <<still in radians.

With R and the the chord length, a, 2 sides of the right triangle are known so pythagorus can tell us the length of the remaining side, r.

r=sqrt(R^2-(a/2)^2)

The relationship between R, r, and h is given by the equation h=R-r

OK, Furikan, I'm convinced--you really can determine theta given a and s. Cool! And thence, you've derived the rest. Neat! Thank you.

[Prof. Templeton]

Good job! Orykle comes in with the closest answer first, and Furikan has the closest answer to mine plus a decent explanation.

60.75 meters for the length of the sagitta (height between chord and arc). The 60 degree temperature rise causes the 5km pipe to expand almost 2 meters. But the ends are fixed so the pipe would bulge upward almost 61 meters.

I'll put up a detailed explanation when I get some more time.

[Guest]

Good job! Orykle comes in with the closest answer first, and Furikan has the closest answer to mine plus a decent explanation.

60.75 meters for the length of the sagitta (height between chord and arc). The 60 degree temperature rise causes the 5km pipe to expand almost 2 meters. But the ends are fixed so the pipe would bulge upward almost 61 meters.

I'll put up a detailed explanation when I get some more time.

Consider the excellent diagram that Furikan made:

http://brainden.com/forum/uploads/monthly_03_2009/post-14039-1236807711.gif

You are saying that (with rounding)

a = 5000M

s= 5002M

h = 60M

No way.

[HoustonHokie]

Consider the excellent diagram that Furikan made:

http://brainden.com/forum/uploads/monthly_03_2009/post-14039-1236807711.gif

You are saying that (with rounding)

a = 5000M

s= 5002M

h = 60M

No way.

It really is. The radius of the arc segment is really large - I get about 50.8 km. If you cut a 5 km chord across a 50 km circle, the distance between the two at the midpoint is 0.06 km, or 60 m. Keep in mind the bulge height is only about 1% of the pipe length.

Of course, in real life, the pipe would have squirmed around like a snake before it finally split at the joints - oil everywhere.

[bonanova]

I spent a while on this and opted out for a rough approximation.

Assume the pipe buckled not in a circular arc, but in two straight pieces.

Since the expansion was a small fraction, this might over-estimate the arc height by 10-15%.

The angle the pipe makes with the ground at each end is theta = cos^-1[5000/5001.968].

The height at the midpoint is then 2500 x sin theta = 70.15 meters.

Subtracting 10-15% gives a range of 59.6m to 63.1m

That served as a sanity check on my exact analysis.

I went wrong somewhere, getting a result of 1768m.

Should be evident my training was in engineering, not math.

Actually it was both.

But never underestimate the value of quick and dirty.

This puzzle is a classic; sometimes named the Railroad Problem.

The exact solution involves transcendental equations, easily solved now by computer,

but close approximations can be made for an algebraic solution.

[Guest]

Good job! Orykle comes in with the closest answer first, and Furikan has the closest answer to mine plus a decent explanation.

60.75 meters for the length of the sagitta (height between chord and arc). The 60 degree temperature rise causes the 5km pipe to expand almost 2 meters. But the ends are fixed so the pipe would bulge upward almost 61 meters.

I'll put up a detailed explanation when I get some more time.

I got 60.74896m

The trick is that one has to solve the problem numerically, specifically, if c is the chord (5000m) and a is the arclength (5001.968m) then a/s = (sinx/x)

where x is half the angle of the sector, which we will call theta. the radius is (a/theta), the apothem, which we will call 'p' is p = rcos(x), and the height is h = r-d

r = 51471.587 m

theta (rad) = 0.097179

theta (deg) = 5.568123

p = 51410.84 m

From Dr. Math on mathforum.org:

The equation f(x) = 0 can be solved numerically using (Sir Isaac) Newton's Method as follows.

Guess a starting value x(0). This will depend heavily on the form of f(x). Then for each n = 0, 1, 2, ..., compute

x(n+1) = x(n) - f[x(n)]/f'[x(n)],

where f'(x) is the derivative of f(x) with respect to x. Continue this until |x(n+1)-x(n)| is smaller than the accuracy sought. Then x(n+1) agrees with the actual root to at least that level of accuracy.

If the starting value for x(0) is close enough to a root, this will converge, and very rapidly. If x(0) is rather far from a root, this process may diverge, or exhibit other undesirable behavior.

To solve, for example, sin(x)/x = k for some constant k > 0 (Case 1 above), is the same as finding a root of the equation

f(x) = sin(x) - kx = 0.

This can be done using Newton's Method as follows.

Guess a starting value of

x(0) = sqrt(6-6k).

Then for each n = 0, 1, 2, ..., compute

x(n+1) = x(n) - (sin[x(n)]-kx(n))/(cos[x(n)]-k).

Example: Solve sin(x)/x = 5000/5001.968 to five decimal places of accuracy. We carry ten places of accuracy in our calculations:

n ----------- x(n) -------------- sin(x[n]) -----------cos(x[n]) --------- sin(x[n])/x(n)

========================================================

0 ------- 0.048586735 ---- 0.048567621 ---- 0.998819897 ---- 0.999606601

1 ------- 0.048589603 ---- 0.048570486 ---- 0.998819757 ---- 0.999606555

2 ------- 0.048589603 ---- 0.048570486 ---- 0.998819757 ---- 0.999606555

Edited March 13, 2009 by Voron X.

[Guest]

it's 2.5 km. if it's a perfect circular ark, then it's half the circle, so all you have to do is divide the length in two to get the radius.

[bonanova]

it's 2.5 km. if it's a perfect circular ark, then it's half the circle, so all you have to do is divide the length in two to get the radius.

Hi UBUH,

Sometimes an OP gives you some red herrings, and you get right to the answer by ignoring them.

In this case, you ignored the expansion coefficient and the temperature change.

That is a critical part of the puzzle, because it tells you how much longer the pipe became.

You made the assumption that an arc is a semicircle and solved that case.

But if it were, the pipe must have grown by a factor pi/2 - which does not agree with the information you ignored.

Here's the place you went wrong ... any portion of a circle is an arc.

Also, it's a good idea to put your solutions into a spoiler.

That way members can choose whether to read it before solving it themselves.