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A friend at work posed this math problem to me the other day. It isn't realistic, but it's still interesting  particularly because answer is surprising.
A steel company has developed an experimental railroad rail. They can make very long rails without joints so they fabricate one that is 1 mile long (5280 feet). They lay it out straight and fix the endpoints to the ground. A few hours later, the rail had expanded in the midday heat. This expansion has added 1 foot of length to the rail, so now it is 5281 feet long. The end points are still attached to the ground, so the rail has warped and is no longer straight.
Assume that the rail bends in a circular arc. The question is "how much has the rail warped at the mid point?" In other words, after adding 1 ft of length, the whole rail has bent. How far is the mid point of bent rail from it's original position?
Without doing any calculations, guess the answer.
Can you come up with a rough approximation that can be computed on a 4banger?
Can you find the exact (within 2 decimal places) answer?
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Very interesting problem. Before doing any math, I guessed maybe .1 inch. So I played with the math, circles, circumference, diameter, radius, and calculated the answer of 3.84 inches, but I have no idea if my assumptions are correct so I wouldn't bet any money on that answer.
I'm looking forward to some answers from those engineers who know much more than I.
Thanks for the challenge on a snowy day in Louisville.
Don
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Off the top of my head, I think that this is like the "string around the earth" problem. My guess is that it's going to be really high in the middle......
While not at all accurate: figure this is approximately a right triangle with the hypotenuse being 1/2 of the new length (2640.5) and the long side being 1/2 of the original length (2640). That makes the height 51+ feet.
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I did guess less than a foot. Calculating with good ol' Pythagoras I get 51.4 feet. Surprise!
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First order approximation according to good ol' Taylor results in SQRT(2640) = 51.4 as well.
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Quote:
Can you find the exact (within 2 decimal places) answer?
44.52 feet.
Yes, I was surprised at how much.
Katie's estimate wasn't too bad.
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Martin, I get 44.4985 on my 50g. I get 78,335.0805ft for the radius of the circle. One of us is probably having a rounding error.
I'll wait a day or two before posting details of the solution. I too was surprised at how big the deflection is. I was also surprised at how far off the "assume a right triangle" approximation is.
Dave
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I've found similar results using the HP33s Solver:
R*SIN(2640.5/R)=2640 > R=
78,335.0803155
ENTER ENTER 2640 x<>y / ASIN COS *  > 44.4984551
Gerson.
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For those with a 42S, a general inverter:
00 { 42Byte Prgm }
01*LBL "INV"
02 FC? 45
03 GTO 00
04 RCL "X"
05 XEQ IND "A"
06 RCL "Y"
07 RTN
08*LBL 00
09 ASTO "A"
10 PGMSLV "INV"
11 STO "Y"
12 CLX
13 STO "X"
14 1
15 SOLVE "X"
16 END
The fact that it's 42 bytes is a coincidence ;)
usage:
ALPHA: name of function to invert (F)
X: value Y for which F(X) = Y
XEQ "INV" > X
eg. to find the value of X for which SINC(X) = 5280/5281, do:
1. define the function SINC(x) = SIN(x)/x for x#0, = 1 for x=0
01*LBL "SINC"
02 X=0?
03 GTO 00
04 SIN
05 LASTX
06 /
07 RTN
08*LBL 00
09 SIGN
10 END
2.A: "SINC"
X: 5280 5281 /
XEQ "INV" > 3.37077587945e2
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I get 44.50 feet (pretty close to Martin's answer).
This is why you need expansion joints in concrete sidewalks and roads. Otherwise, a bit of thermal expansion will cause considerable heaving.
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Quote: This is why you need expansion joints in concrete sidewalks and roads. Otherwise, a bit of thermal expansion will cause considerable heaving.
It is my understanding that those extremely long railway rails basically aren't allowed to expand, i.e. they are anchored firmly enough to prevent them from buckling, and when they heat up or cool down, away from their "nominal" temperature, they just become compressed or stretched, respectively, but they don't move. Anyone who actually knows modern railway engineering firsthand, please correct me. :)
Edited: 21 Jan 2011, 12:01 a.m.
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(While I was busily writing, David Hayden, Gerson, and Dave Shaffer were posting identical results using what appears to be the same approach...)
I'd say that Martin's answer is correct, or very nearly so. I got 44.4984551 feet on my HP32SII.
At a family gathering in the 1970's, I remember my (nonrailroad) engineer grandfather reading this problem and giving the answer for a straight expanded track. He, too, found the height of about 50 feet to be greater than expected. It really illustrates the importance of expansion joints in bridges, lest severe buckling occur during heat waves.
The trigonometric solution for the straight expanded track is simple:
0.5 * sqrt(5281^{2}  5280^{2}) = 51.3833630665
Solving for the circulararc expanded track proved to be more complicated. I got the answer from iterative solution of the wellknown sinc function [sin(x) / x].
(For a graphical illustration, please refer to Thomas Klemm's post deeper in this thread. Notational conversion: A/2 = x; h = e; d = L1)
Define the following:
L is the total length of the arc of expanded track
d is the original length of straight track
r is the radius of the circular arc of track
A is the angle of the full arc [radians]
y is the vertical distance from the center of the arc circle to the track surface
Then,
1. L/2 = r * A/2
2. d/2 = r * sin (A/2)
3. d/2 = y * tan (A/2)
Some algebra using equations 1 and 2 yields
sin (A/2) d
 = 
(A/2) L
With d = 5280 ft and L = 5281 ft,
SOLVE on the HP32SII and HP42S yields
A/2 = 0.0337077587945 rad
The rise of the track (h) = r  y
From equations 1 and 3,
L/2 d/2
h =   
A/2 tan (A/2)
h = 44.4984551
Alternatively from equations 2 and 3,
[ 1 1 ]
h = d/2 *     
[ sin (A/2) tan (A/2) ]
h = 44.49845532
The HP15C, HP34C, and HP41/Advantage produce the following results:
A/2 = 0.03370777144 rad
h = 44.49846
Gerson stated below that the answer correct to 12 digits is
h = 44.4984550191
which is a bit lower than either of my results.
The differences are explained by lastdigit roundoff error.
Addendum
WolframAlpha can be utilized to solve many digits of an equation.
Formula:
sinc(x) = 5280/5281
==> x = 0.0337077588098794293482184983
This value can be used in Windows calculator to calculate the rise (h) to 24 digits or so:
2640.5 2640
h =   
x tan(x)
h = 44.49845501910079925455416...
Edited: 30 Jan 2011, 5:51 p.m. after one or more responses were posted
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Quote:
It really illustrates the importance of expansion joints in bridges, lest severe buckling occur during heat waves.
Another place where expansion has to be taken into account is piping systems. We do hot/chilled water piping design and are very careful to allow for the expasion/contraction that can take place in the system. In underground piping systems, concrete thrust blocks are added at bends to ensure that the expansion takes place in the expansion joints and not have the pipe move at the direction change. It's not uncommon to have miles of underground piping on a large campus environment.
But I have to admit, I was somewhat supprised at the 50 feet expansion over one mile.
Bill
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Quote:
But I have to admit, I was somewhat surprised at the 50 feet expansion over one mile.
To be technically precise, the lateral displacement was (roughly) 50 ft. The expansion was 1 ft. Incidentally, this corresponds to about 23 degrees F temperature rise.
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Quote:
The expansion was 1 ft. Incidentally, this corresponds to about 23 degrees F temperature rise.
Thanks, Martin, I was just going to check this out. Here in my area (southern Brazil), 40 °F temperature rise or greater is not uncommon.
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Quote:
lateral displacement was (roughly) 50 ft.
You're right  Expansion is only 1 foot.
Thanks,
Bill
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Karl,
Quote:
(While I was busily writing, David Hayden, Gerson, Dave Shaffer were posting identical results using what appears to be the same approach...)
I don't remember the word you used in your first version of this text in place of busily. It was not diligently, but it had a close meaning and expressed better what you did. Congratulations for the nice work and your detailed and didactic explanations, as always. It takes time to provide a complete solution as you have done. Most here just don't have time to, anyway it is a good thing they occasionally take the time to solve problems and post the results so we can compare them to our own. As of me, I am just kind of lazy.
Quote:
I'd say that Martin's answer is correct, or very nearly so. I got 44.4984551 feet on my HP32SII.
I do agree. Martin is a Civil Engineer, if I am not wrong, so his twodigit answer is good enough for practical purposes. I remember myself solving transmission lines problems (involving grossbeak aluminum cables and other birds names) and the answers to six decimal places during examinations, when two would have sufficed. My professor found that strange but would never say anything :)
Quote:
Solving for the circulararc expanded track proved to be more complicated.
I am glad you have said this. So, I am not the only one :)
Here is my solution, using a similar approach. Yesterday, at first I had used an expression involving the ASIN function at first, then switched to SIN. I have decided to use ASIN again, only because it's similar to the answer I got through Calculus. Anyway, both are equivalent:
yaxis

r___
e ^  L L = half length of the arc (2,640.5 ft)
v ______\ r = radius of the circular arc
^  d /
 / h = height from cente of the arc to the track
 / e = r  h (vertical expansion)
h  /r
A / A = angle in radians
 /
/
v _______________ xaxis
0 r
< d > d = half lenght of track (2,640 ft)
Using
2L = r*2A (length of arc subtended by central arc)
and
d
sin(A) = 
r
L
We get A = 
r
/ d \
and A = arcsin  
\ r /
which implies
/ d \
L = r*arcsin  
\ r /
Replacing L and d with their known values, 2640.5 and 2640, respectively, we get
r*arcsin(2640/r) = 2640.5
when solved for r (initial estimate r = 50,000) on the HP32SII in radians mode yieds
r = 78,335.0808851 ft
We are now able to compute the vertical expansion e:
From the drawing (well, sort of),
h = sqrt(r^2  d^2)
h = sqrt(78,335.0808851^2  2640^2)
h = 78,290.5824303 ft
e = r  h
e = 78,335.0808851 ft  78,290.5824303 ft
e = 44.4984548 ft
This morning, hoping to avoid the solver, I tried Calculus. It turns out I got the same equation:
From the Length of an Arc Formula:
L = Integrate(x=a, b, sqrt(1+[f'(x)]^2)dx)
(SCHWARTZ Abraham, Calculus and Analytic Geometry. Holt, Rinehart and Winston, 3rd edition)
f(x) = sqrt(r^2  x^2)
f'(x) = x*sqrt(r^2  x^2)/(x^2  r^2)
[f'(x)]^2 = x^2/(x^2r^2)
L = Integrate(x=O, 2640, sqrt(1  x^2/(x^2  r^2)))
L = r*arcsin(2640/r)
or r*arcsin(2640/r) = 2640.5
On the HP50g:
'sqrt(R^2X^2' DERVX x^2 1 + sqrt > 'ABS(R)/ABS(R^2X^2)*SQRT(R^2X^2)'
INTVX > 'ASIN(X/R)*(R*SIGN(R)/SIGN(R^2X^2))'
Since R > 0 and X=2640 > 0, this simplifies to
'R*ASIN(X/R)')
Regards,
Gerson.
Edited: 21 Jan 2011, 1:26 p.m.
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Gerson's first solution is exactly the way that I solved it. Thanks for the great explanation and the clever ASCII diagram.
Dave
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Gerson 
Thank you for the compliments. My only objective was to solve the problem with minimal effort in a straightforward fashion using the 'favored' tools we have at our disposal, and to show the procedure clearly. I believe that I'd examined a problem like this one before, but could not readlity deduce any apparent trigonometric formula to solve it.
The mathematical work done by you and Thomas Klemm in particular was more impressive, and Thomas also prepared a nice diagram to accompany his symbolic algebra.
Of course, the algebraic methods of estimation worked only because the angle was very small. In the days before calculators and computers, I suppose that an engineer or architect would refer to precalculated tables of the 'sinc' function to solve it. The function is used in signal processing.
 Karl
Edited: 23 Jan 2011, 1:45 p.m. after one or more responses were posted
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Thanks, Karl, for your comments. As of the sinc function, I thought of finding an approximation for it to make the solution not dependant of the solver. Sinc(x) ~ cos(x*sqrt(3)/3) might be an option and would have yielded 3 correct decimal places in your method (44.49827). The constant has no mathematical meaning, but works for small angles. That's cheating, however...
Gerson.
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Gerson,
Very nice
approximation.
 Pauli
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That was found when solving SIN(A)/ACOS(A*X)=0 for X, A=3.37077587945e2 (that's A/2 in Karl's post above). I got X=5.7734397696e1 which is very close to sqrt(3)/3 (off by 7.29e6). For X=0.5 up to X=1 the difference ranges from 1.e5 to 6.6e3. I tried a correction term but I discarded it because it was not suitable for the problem (y ~ cos(x*sqrt(3)/3)+(2+sqrt(3))/1000*x^4).
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Quote:
Very nice
approximation.
Paul,
What about this one? Of course, this is not related to the orignal problem anymore.
Gerson.
Edited: 24 Jan 2011, 5:34 p.m.
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Quote:
The constant has no mathematical meaning,
I should have said "the constant has no mathematical meaning I am aware of". Sqrt(3)/3 appears to be the exact value for k in y = sinc(x)  cos(k*x) that yields maximum accuracy. A little more or a little less will degenerate the approximation and make the function either change sign or oscillate around zero. The Wolfram Alpha's graph can show y(x) is always positive until the range is narrowed down to [1e3, 1e3], when the machine precision limit is reached.
By the way, an exact equivalence is
sinc(x) = 1/(Gamma(1 + x/pi)*Gamma(1  x/pi))
(See MathWorld Sinc Function page, formula #10)
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Although a first order approximation, about 50 feet.
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Yeah, as someone mentioned, this problem is similar to the "rope around the world" problem, which was mentioned here.
There, an extra meter in the length of the rope caused it to be raised up ~120 meters over the surface of the earth.
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This problem has been solved a lot, but not a lot has been said about why the answer is so huge for a such a small input, so I want to talk about that.
The reason is that the answer function does not have a Taylor series  at least not at x=0 (if x is the slack in the rope around the world, or the expansion of the railroad).
For the "rope around the world" problem, the first step (in my solution) was to solve
y + 2 tan(y/2) = x/r
taking the taylor series of the left side, that's
y^3/12=x/r
or
y = cuberoot(12*x/r)
Then you plug y into
r*(cos(y/2)+sin(y/2)*tan(y/2)1)
The lowest taylor series there is
r*y^2/8.
So finally, the height is
2^(1/3)*r/4*(3*x/r)^(2/3)
So, the height goes as x^(2/3) power. This has infinite slope at x=0, explaining why the answer grows so rapidly even for very small x.
I also should point out that that first order approximation is accurate enough to agree to all 5 digits of my originally quoted answer.
For the railroad problem, not showing my work (there was way more work, at least for me, than the rope problem), but the final answer to first approximation is
sqrt(6*m*x)/4
with m=5280, and x=1, this gives 44.497
Note: In form, this is very similar to the trigonometric solution, but is much more accurate! The trigonometric solution to first approximation is sqrt(2*m*x)/2 ~ 51.38 for the given values.
Before going through the work myself, I might have thought the first approximation formulas (trigonometric versus circular) would be the same, but they're not.
But again, the answer goes as squareroot of x, which has infinite slope at x=0.
Sort of interesting that the basic idea of the problems is so similar, as well as the large increase in the answer for small values of x, but the power dependency on x for the two problems is different.
Edited: 21 Jan 2011, 6:40 p.m.
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But doesn't it seem like there should be a simple explanation for the factor of sqr(3)/2 difference between the two methods?
If so, that with the trigonometric approach would seem to be the simplest method to solve the problem.
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I think I figured that out, and in the process came up with a slightly different way to solve the problem.
Imagine that we have the normal "small angle" slice of a circle.
Let's just set r=1. The green line is sin(t), the blue line (after a bit of geometry) is sqrt(2(1cos(t))), and the black arc is of course t.
What we want to find out is how the blue line and the black arc compare to the green line when t is small.
The taylor series of sqrt(2(1cos(t)))/sin(t) = 1+t^2/8
The taylor series of t/sin(t)= 1 + t^2/6
So, if the green line is normalized to 1, the blue line (trigonometric solution) is normalized to 1+t^2/8, and the circle is a little bit larger (naturally), 1+t^2/6.
t^2/6 IS the added length (or half of it, but it doesn't really matter since we're normalizing everything)
Since the green line is normalized to 1, that basically means we're dividing by m. The extra length is then x/m.
This means t = sqrt(6x/m)
plugging that into t^2/8 (the extra for the trigonometric case), we get 3x/(4m).
In other words, if we replace the extra length x/m with the "adjusted" extra length 3x/(4m), we can use the trigonometric solution and get the same answer.
The trigonometric solution was m*sqrt(2(x/m))/2.
So the real correct solution is m*sqrt(2 (3x/4m ))/2 = sqrt(6xm)/4.
By the way, I see some people elsewhere in the thread concerned about accuracy for the sinc function. It's probably not worth worrying about. Even this first order solution agrees with the actual solution to within 1% for as much as a 350 foot expansion!!
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Very nice!
But I think we can do without the indirection of the "trigonometric solution", once we realize that the small angle in the greenredblue triangle is in fact t/2.
Now we can use t to calculate the lateral displacement e:
Cheers
Thomas
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Quote:
Can you come up with a rough approximation that can be computed on a 4banger?
Just using the Taylor expansion of the trigonometric functions I got:
Quote:
I was also surprised at how far off the "assume a right triangle" approximation is.
This shows you can do better even without a fancy solver.
Cheers
Thomas
Edited: 22 Jan 2011, 2:35 a.m.
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Okay, so your formula looked like my formula. So why did mine give a slightly better result?
They're not actually the same. Mine was
sqr(3*m*x/8)
while yours was (effectively, since you substituted in x=1)
sqr(3*(m+x)*x)/8)
That's actually not a first order formula, since it has an x^2 term. If m is large compared to x, you drop x in m+x.
You might think going to second order would improve the accuracy, not decrease it, but that might not be the correct second order approx., if second order terms were dropped earlier.
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Quote:
For the railroad problem, not showing my work (there was way more work, at least for me, than the rope problem), but the final answer to first approximation is
sqrt(6*m*x)/4
with m=5280, and x=1, this gives 44.497
Honestly I must admit that I didn't realize that you provided a solution to this problem but was thinking you were only dealing with the rope problem. I'd be interested to see your approach and might not be the only one.
Best regards
Thomas
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The approach is exactly the same in terms of setting up the problem. You just have to be careful when selecting the number of terms to retain.
I went ahead and got the second order solution. It is
y = sqrt(6mx + 9/5 x^2)/4
That gives an answer that agrees with the "exact" answer to at least 8 places.
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Power Series of Equation
Instead of solving the following equation for t I tried to calculate the inverse of the Taylor expansion using WolframAlpha.
As can be seen by the graph there's a minimum at the point (0, 1).
This makes the inversion difficult since the first derivative of the inverse function is infinite. Thus I use the following substitution:
Now the graph has a positive derivative:
This leads to the following equation:
Let's try to find the Taylor series expansion:
Series[Sqrt[x]/Sin[Sqrt[x]], {x, 0, 4}]
Thus we end up with:
Inversion of Power Series
Starting with the equation f(y(x)) = x I calculated little by little the derivations:
D[x=f[y[x]],{x,1}]
D[x=f[y[x]],{x,2}]
D[x=f[y[x]],{x,3}]
D[x=f[y[x]],{x,4}]
From the coefficients of the power series we can easily calculate the derivation at x = 0:
y'[x] = 1/6
y''[x] = 7/180
y'''[x] = 31/2520
y''''[x] = 127/25200
If we put these values into the equations below we get a system of linear equations:
f'[y[x]] y'[x] == 1
y'[x]^2 f''[y[x]] + f'[y[x]] y''[x] == 0
3 y'[x] f''[y[x]] y''[x] + y'[x]^3 f'''[y[x]] + f'[y[x]] y'''[x] == 0
6 y'[x]^2 y''[x] f'''[y[x]] + f''[y[x]] (3 y''[x]^2 + 4 y'[x] y'''[x]) + y'[x]^4 f''''[y[x]] + f'[y[x]] y''''[x] == 0
Solve[{
{1/6, 0, 0, 0},
{7/180, (1/6)^2, 0, 0},
{31/2520, 3*1/6*7/180, (1/6)^3, 0},
{127/25200, 3*(7/180)^2+4*1/6*31/2520, 6*(1/6)^2*7/180, (1/6)^4}
}.{f1,f2,f3,f4}=={1,0,0,0},{f1,f2,f3,f4}]
Here's the solution:
From these values we can finally calculate the coefficients of the inverse power series:
Calculation of the lateral displacement
We use the same formula as before:
Now we plug the expression for t into 4 tan^{2}(t/2) and calculate the power series of that:
Series[4 Tan[Sqrt[6y21y^2/5+564y^3/175459y^4/175]/2]^2,{y,0,4}]
After a last substitution y=d/m we finally get:
Using only the first three terms I get the same value to 12 places as Gerson stated in message #36:
Quote:
e = 44.4984550191
Cheers
Thomas
PS: InverseSeries
and ComposeSeries
don't seem to work with WolframAlpha. That would have made this much easier.
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And this was supposed to be a 5minute challenge...
Quote:
Impressive work! You and Crawl have derived an elegant and compact formula for a nontrivial problem.
Here are the numeric results the formula gives gives out, from 1st through 4th order:
1st: 44.497190922573977692
2nd: 44.498455029360289016
3rd: 44.498455019099827724
4rd: 44.498455019100799359
actual: 44.498455019100799255
The latter agrees with the actual result to 17 significant digits!
In order to get all those digits, I had to ask WolframAlpha
for a numeric solution (the limit of my skills). First "solve r*arcsin(5280/(2*r))=5281/2 for r", then using the numeric value for r thus obtained: "78335.08050455416533561551559  sqrt(78335.08050455416533561551559^2  5280^2/4)".
Regards,
Gerson.
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Meanwhile I had access to Mathematica where the whole thing is much easier:
4 Tan[Sqrt[InverseSeries[Series[Sqrt[x]/Sin[Sqrt[x]]  1, {x, 0, 6}]]]/2]^2
Funny to see how the coefficients start to become very complicated.
Just out of curiosity: Is it possible to do something like this with the CAS of the HP50g?
Cheers
Thomas
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Quote:
Quote:
Can you come up with a rough approximation that can be computed on a 4banger?
Actually, your solution meets David Hayden's second question ("Can you find the exact (within 2 decimal places) answer?") while being able to do it on a 4banger. Well done!
Quote:
Quote:
I was also surprised at how far off the "assume a right triangle" approximation is.
This shows you can do better even without a fancy solver.
By resorting to Pythagoras instead to the Taylor's series for cosine, the approximation is improved:
e = r  sqrt(r^{2}  ((L1)/2)^{2})
e = L*sqrt(L/24)  sqrt(L^{3}/24  (L1)^{2}/4)
e = 44.4971906
That's very close to Crawl's approximation but, unlike his or yours, far from concise.
Adding another term to the first series would require solving a quartic equation in order to find the radius. Fortunately, it is reducible to a quadratic equation and can be done by hand. Eventually I got, following your steps:
r = L/(2*sqrt(10*(1sqrt(16/(5*L)))))
which, when plugged into the first equation above, yields
e = 44.498455 (44.49845 on the HP12C and probably even less accurate on an ordinary 4banger)
As a comparison, the actual result to 12 significant digits is
e = 44.4984550191
Regards,
Gerson.
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Your method is more accurate than you realise. If you solve for the radius as you suggest, the radius is 78315.056. Next solve directly for the angle 2x, by arc/radius. ie 5280/78315.056. This is .067420 radians and equivalent to 3.862881 degrees. Therefore x = 1.931441 degrees. And (1 Cosx) = .000568
And finally defelection = .000568*78315.056 = 44.492977
This answer is only 0.005 different to the correct result. If you work backwards from the above the chord length is 5279.00051. So for all practical purposes the above is correct.
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Through a coincidence worthy of a Hollywood movie, I actually had a somewhat practical application of this very problem today.
We have some chairs on our porch. The seats are supported by an elastic webbing similar to this. The original webbing stretched out over the years so I cut it, but it just stretched out more and I decided to replace it.
A little hunting on the internet led me to elastbelt webbing which stretches about 7%10%.
When you install the webbing, you pull it tight, so say it can only stretch an additional 5% once installed. If the seat is 20" wide and the webbing stretches 5%, how much will it sag when I sit in it? Will I be able to get out of the chair to say hello to the neighbors when they walk by? Will it sag enough to be comfortable?
I may have done the calculations wrong, but I figured it should sag about 2.7590"  Perfect! :)
