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Five-minute challenge
Message #1 Posted by David Hayden on 20 Jan 2011, 2:00 p.m.

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 4-banger?
Can you find the exact (within 2 decimal places) answer?

      
Re: Five-minute challenge
Message #2 Posted by Don Shepherd on 20 Jan 2011, 3:31 p.m.,
in response to message #1 by David Hayden

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

            
Re: Five-minute challenge
Message #3 Posted by Katie Wasserman on 20 Jan 2011, 4:43 p.m.,
in response to message #2 by Don Shepherd

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.

                  
Re: Five-minute challenge
Message #4 Posted by Walter B on 20 Jan 2011, 5:15 p.m.,
in response to message #3 by Katie Wasserman

I did guess less than a foot. Calculating with good ol' Pythagoras I get 51.4 feet. Surprise!

                        
Re: Five-minute challenge
Message #5 Posted by Walter B on 20 Jan 2011, 5:26 p.m.,
in response to message #4 by Walter B

First order approximation according to good ol' Taylor results in SQRT(2640) = 51.4 as well.

      
Re: Five-minute challenge
Message #6 Posted by Martin Pinckney on 20 Jan 2011, 4:46 p.m.,
in response to message #1 by David Hayden

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.

            
Re: Five-minute challenge
Message #7 Posted by David Hayden on 20 Jan 2011, 10:09 p.m.,
in response to message #6 by Martin Pinckney

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

                  
Re: Five-minute challenge
Message #8 Posted by Gerson W. Barbosa on 20 Jan 2011, 10:44 p.m.,
in response to message #7 by David Hayden

I've found similar results using the HP-33s Solver:

R*SIN(2640.5/R)=2640                    ->     R=
                                               78,335.0803155

ENTER ENTER 2640 x<>y / ASIN COS * - -> 44.4984551

Gerson.

                        
Re: Five-minute challenge
Message #9 Posted by Werner on 25 Jan 2011, 5:49 a.m.,
in response to message #8 by Gerson W. Barbosa

For those with a 42S, a general inverter:

00 { 42-Byte 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" -> 3.37077587945e-2

            
Re: Five-minute challenge
Message #10 Posted by Dave Shaffer (Arizona) on 20 Jan 2011, 10:10 p.m.,
in response to message #6 by Martin Pinckney

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.

                  
Re: Five-minute challenge
Message #11 Posted by Thomas Okken on 20 Jan 2011, 11:56 p.m.,
in response to message #10 by Dave Shaffer (Arizona)

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.

                        
Re: Five-minute challenge
Message #12 Posted by Paul Gaster on 21 Jan 2011, 12:35 a.m.,
in response to message #11 by Thomas Okken

Sounds like some rails designs have gaps and others are just restrained from moving like you said. See here:

http://en.wikipedia.org/wiki/Track_(rail_transport)

            
Re: Five-minute challenge
Message #13 Posted by Karl Schneider on 20 Jan 2011, 11:13 p.m.,
in response to message #6 by Martin Pinckney

(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 HP-32SII.

At a family gathering in the 1970's, I remember my (non-railroad) 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(52812 - 52802) = 51.3833630665

Solving for the circular-arc expanded track proved to be more complicated. I got the answer from iterative solution of the well-known
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 = L-1)

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 HP-32SII and HP-42S 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 HP-15C, HP-34C, and HP-41/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 last-digit 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

                  
Re: Five-minute challenge
Message #14 Posted by Bill (Smithville, NJ) on 21 Jan 2011, 6:06 a.m.,
in response to message #13 by Karl Schneider

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

                        
Re: Five-minute challenge
Message #15 Posted by Martin Pinckney on 21 Jan 2011, 11:31 a.m.,
in response to message #14 by Bill (Smithville, NJ)

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.
                              
Re: Five-minute challenge
Message #16 Posted by Gerson W. Barbosa on 21 Jan 2011, 12:31 p.m.,
in response to message #15 by Martin Pinckney

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.

                              
Re: Five-minute challenge
Message #17 Posted by Bill (Smithville, NJ) on 21 Jan 2011, 1:03 p.m.,
in response to message #15 by Martin Pinckney

Quote:
lateral displacement was (roughly) 50 ft.

You're right - Expansion is only 1 foot.

Thanks,

Bill

                  
Re: Five-minute challenge
Message #18 Posted by Gerson W. Barbosa on 21 Jan 2011, 12:25 p.m.,
in response to message #13 by Karl Schneider

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 HP-32SII.

I do agree. Martin is a Civil Engineer, if I am not wrong, so his two-digit 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 circular-arc 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:

   y-axis

| 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 _|___________|___ x-axis 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 HP-32SII 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^2-r^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 HP-50g:

'sqrt(R^2-X^2' DERVX x^2 1 + sqrt --> 'ABS(R)/ABS(R^2-X^2)*SQRT(R^2-X^2)'

INTVX --> 'ASIN(X/R)*(R*SIGN(R)/SIGN(R^2-X^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.

                        
Re: Five-minute challenge
Message #19 Posted by David Hayden on 21 Jan 2011, 3:38 p.m.,
in response to message #18 by Gerson W. Barbosa

Gerson's first solution is exactly the way that I solved it. Thanks for the great explanation and the clever ASCII diagram.

Dave

                        
Re: Five-minute challenge
Message #20 Posted by Karl Schneider on 22 Jan 2011, 5:41 p.m.,
in response to message #18 by Gerson W. Barbosa

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 pre-calculated 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

                              
Re: Five-minute challenge
Message #21 Posted by Gerson W. Barbosa on 22 Jan 2011, 9:04 p.m.,
in response to message #20 by Karl Schneider

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.

                                    
Re: Five-minute challenge
Message #22 Posted by Paul Dale on 22 Jan 2011, 9:28 p.m.,
in response to message #21 by Gerson W. Barbosa

Gerson,

Very nice approximation.

- Pauli

                                          
Re: Five-minute challenge
Message #23 Posted by Gerson W. Barbosa on 22 Jan 2011, 10:49 p.m.,
in response to message #22 by Paul Dale

That was found when solving SIN(A)/A-COS(A*X)=0 for X, A=3.37077587945e-2 (that's A/2 in Karl's post above). I got X=5.7734397696e-1 which is very close to sqrt(3)/3 (off by 7.29e-6). For X=0.5 up to X=1 the difference ranges from 1.e-5 to 6.6e-3. 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).

                                          
Re: Five-minute challenge
Message #24 Posted by Gerson W. Barbosa on 24 Jan 2011, 4:41 p.m.,
in response to message #22 by Paul Dale

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.

                                    
Re: Five-minute challenge
Message #25 Posted by Gerson W. Barbosa on 23 Jan 2011, 5:34 p.m.,
in response to message #21 by Gerson W. Barbosa

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 [-1e-3, 1e-3], 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)

      
Re: Five-minute challenge
Message #26 Posted by Thomas Chrapkiewicz on 20 Jan 2011, 5:09 p.m.,
in response to message #1 by David Hayden

Although a first order approximation, about 50 feet.

      
Re: Five-minute challenge
Message #27 Posted by Crawl on 21 Jan 2011, 12:42 p.m.,
in response to message #1 by David Hayden

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.

            
Re: Five-minute challenge
Message #28 Posted by Crawl on 21 Jan 2011, 6:38 p.m.,
in response to message #27 by Crawl

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.

                  
Re: Five-minute challenge
Message #29 Posted by Crawl on 21 Jan 2011, 7:08 p.m.,
in response to message #28 by Crawl

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.

                        
Re: Five-minute challenge
Message #30 Posted by Crawl on 24 Jan 2011, 1:53 p.m.,
in response to message #29 by Crawl

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(1-cos(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(1-cos(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!!

                              
Re: Five-minute challenge
Message #31 Posted by Thomas Klemm on 24 Jan 2011, 10:05 p.m.,
in response to message #30 by Crawl

Very nice!

But I think we can do without the indirection of the "trigonometric solution", once we realize that the small angle in the green-red-blue triangle is in fact t/2.

Now we can use t to calculate the lateral displacement e:

Cheers
Thomas

      
Re: Five-minute challenge
Message #32 Posted by Thomas Klemm on 22 Jan 2011, 1:12 a.m.,
in response to message #1 by David Hayden

Quote:
Can you come up with a rough approximation that can be computed on a 4-banger?

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.

            
Re: Five-minute challenge
Message #33 Posted by Crawl on 22 Jan 2011, 8:09 a.m.,
in response to message #32 by Thomas Klemm

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.

                  
Re: Five-minute challenge
Message #34 Posted by Thomas Klemm on 22 Jan 2011, 9:38 a.m.,
in response to message #33 by Crawl

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

                        
Re: Five-minute challenge
Message #35 Posted by Crawl on 22 Jan 2011, 11:16 a.m.,
in response to message #34 by Thomas Klemm

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.

                              
Re: Five-minute challenge
Message #36 Posted by Thomas Klemm on 26 Jan 2011, 10:30 a.m.,
in response to message #35 by Crawl

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 tan2(t/2) and calculate the power series of that:

Series[4 Tan[Sqrt[6y-21y^2/5+564y^3/175-459y^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.

                                    
Re: Five-minute challenge
Message #37 Posted by Gerson W. Barbosa on 26 Jan 2011, 8:44 p.m.,
in response to message #36 by Thomas Klemm

And this was supposed to be a 5-minute challenge...

Quote:

Impressive work! You and Crawl have derived an elegant and compact formula for a non-trivial 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.

                                          
Re: Five-minute challenge
Message #38 Posted by Thomas Klemm on 27 Jan 2011, 9:16 a.m.,
in response to message #37 by Gerson W. Barbosa

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 HP-50g?

Cheers
Thomas

                                                
Re: Five-minute challenge
Message #39 Posted by Gerson W. Barbosa on 31 Jan 2011, 11:50 a.m.,
in response to message #38 by Thomas Klemm

Quote:
Just out of curiosity: Is it possible to do something like this with the CAS of the HP-50g?

I fear the HP-50g lacks InverseSeries, otherwise I think it would handle that.

By the way, I've just found someone else's attempt:

http://www.moderatenerd.com/2009/12/15/the-expanding-rail-problem/

Cheers,

Gerson.

                                                      
Re: Five-minute challenge
Message #40 Posted by Thomas Klemm on 1 Feb 2011, 4:05 p.m.,
in response to message #39 by Gerson W. Barbosa

Thanks for both the answer and the link. Looks promising. It seems there's nothing new under the sun and Google has an answer to every question.

Cheers
Thomas

                                                
Re: Five-minute challenge
Message #41 Posted by Gerson W. Barbosa on 31 Jan 2011, 1:53 p.m.,
in response to message #38 by Thomas Klemm

Quote:
Funny to see how the coefficients start to become very complicated.

Thomas,

The sixth term simplifies to 683073/50050000*x^6:

http://www.wolframalpha.com/input/?i=4*%28sqrt%286%29*%289590497281*sqrt%283%2F2%29%2F44844800000%2B1%2F6*%28-991213011*sqrt%283%2F2%29%2F1379840000-93447*sqrt%286%29%2F336875%29%29-3147579%2F1971200000%29

It would be interesting if there was a way to force the first terms to be left unsimplified, to see how they generalize (if they do at all).

Cheers,

Gerson.

            
Re: Five-minute challenge
Message #42 Posted by Gerson W. Barbosa on 22 Jan 2011, 10:32 a.m.,
in response to message #32 by Thomas Klemm

Quote:
Quote:
Can you come up with a rough approximation that can be computed on a 4-banger?

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 4-banger. 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(r2 - ((L-1)/2)2)

e = L*sqrt(L/24) - sqrt(L3/24 - (L-1)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*(1-sqrt(1-6/(5*L)))))

which, when plugged into the first equation above, yields

e = 44.498455 (44.49845 on the HP-12C and probably even less accurate on an ordinary 4-banger)

As a comparison, the actual result to 12 significant digits is

e = 44.4984550191

Regards,

Gerson.

            
Re: Five-minute challenge
Message #43 Posted by Stuart Sprott on 24 Jan 2011, 7:34 a.m.,
in response to message #32 by Thomas Klemm

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.

      
Re: Five-minute challenge
Message #44 Posted by David Hayden on 29 Jan 2011, 12:16 a.m.,
in response to message #1 by David Hayden

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! :)


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