# numerical computation contest

*A diversion, for anyone so inclined.

*

Does the answer have to be completely numerical, or can it contain known variables/equations (for example, could I say pi/8 versus 0.392699…, or 2*sin (1) were either of those the actual answer. Disclaimer: I have done no math at this point, and I highly doubt that I randomly picked the answer :P)

Looking for a numerical answer, in decimal form. Winner is person who gets answer correct to the most decimal digits.

edit:
If you can find an exact explicit closed-form analytical solution using only add, subtract, multiply, divide, powers, roots, exponentiation, logarithms, trig functions, and inverse trig functions, you will be declared the winner.
*
*

I believe the answer is 44.4984550191007992545541 feet

(although I used a “wolfram alpha method” , not an analytic one to solve the equation)

Is this close?

h = (in ft)
44.498455019100799254554160016733598949968964352229561211688059554658168699956051975334723610652925014422691969000110577445163659978002373921158891281961079122339560352790946889732641412850383528339799280892308107161666785352632556370781259728977129034979416624799668990733212350734392869697893501318891119435591779774054312719121214164103632340254095275303249287104781014962478813559412501235222884006058879524106347465534347539833791041297948729285690841569378338955156536796873312602070842136063260

The arc is only one foot longer than the chord. How could the answer possibly be larger than one?

EDIT: Anyway, here is my work. SPOILERS!
/http://i.imgur.com/2IYR8.png

So my guess is zero. It’s not exact, but close.

Close enough. Tell the folks how you did it.

Create a simple approximation. Imagine a point C that lies at the top of the line H. You can then create a triangle ABC, where length AB is known (5280 ft) and length BC = AC = 5281/2 = 2640.5.

We know that h bisects line AB - lets call the intersection between AB and h to be H. we know that the length AH = BH = 5280/2 = 2640.

So, now we know two sides to the right triangle AHC - AH and AC. Taking the square of the hypotenuse minus the square of one side gives us the square of the other side. In other words, 2640.5^2 - 2640^2 = h^2 (the pythagorean theorem).

So, in this *extremely *rough approximation, we get h = 51.383. Given that, it’s not hard to imagine that Christopher’s answer could be correct, to some number of decimal places.

Thanks!

498 decimal places

I checked them. They appear to be correct.

Since this is a circular arc, I start with a circle. Based off of a diagram like this, I draw a triangle and take theta to be as is shown in that picture. From the right triangle, we know radius * sin(theta) = 5280 / 2. From the definition of arc length, we also know that radius * 2 * theta = 5281. We have two equations and two unknowns, so I substituted one into the other and let Mathematica determine theta. From theta, we easily get the radius. Next, we use this equation relating chord length and the height to the radius. Again, I let Mathematica handle the equation to a ridiculous number of decimal places.

Mathematica Code

``````
Clear[t, h, r, c, a]
c = 5280
a = 5281
t = t /.
FindRoot[a/(2 t) * Sin[t] == c/2, {t, 1.8},
AccuracyGoal -> 100000, PrecisionGoal -> 100000,
WorkingPrecision -> 500]
r = a/(2 t)
FindRoot[r == c^2/(8 h) + h/2, {h, 100}, AccuracyGoal -> 100000,
PrecisionGoal -> 100000, WorkingPrecision -> 500]
``````

EDIT: I can get more decimal places …

I was bored and tried to find a closed form by hand, it didn’t work out by the time I refreshed it and it was over

Here’s Maxima code:

``````fpprec: 600\$
L:5280\$ eps:1\$
y: x/sin(x)-(L+eps)/L\$
a: bf_find_root(y,x,1/1000,4/100);
h: (L/2)*(1-cos(a))/sin(a);
``````

EDIT: I can get more decimal places …

So can I

``````Maxima 5.27.0 http://maxima.sourceforge.net
using Lisp GNU Common Lisp (GCL) GCL 2.6.8 (a.k.a. GCL)
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.

(%i1) fpprec: 600\$
L:5280\$ eps:1\$
y: x/sin(x)-(L+eps)/L\$
a: bf_find_root(y,x,1/1000,4/100);
h: (L/2)*(1-cos(a))/sin(a);

4.4498455019100799254554160016733598949968964352229561211688059554658168\
699956051975334723610652925014422691969000110577445163659978002373921158891281\
961079122339560352790946889732641412850383528339799280892308107161666785352632\
556370781259728977129034979416624799668990733212350734392869697893501318891119\
435591779774054312719121214164103632340254095275303249287104781014962478813559\
412501235222884006058879524106347465534347539833791041297948729285690841569378\
338955156536796873312602070842136063260258182278201500825791387457906343098887\
9258590281352783951539842610751922916884732140489864979154825b1
(%i7)
``````

4.4498 … I thought it was 44.498 …

Um… The exponent is at the end of those 600 digits

Russ,

I understand you intended this one as a programming exercise – hence you posted it in the programming forum.

However, it can also serve as an example of engineering approximation. Jon showed earlier how the Pythagorean theorem gives 51.48 ft as a rough approximation.

For a slightly better one, I used the Taylor series expansion for the sinc function, defined as sinc(x) = sin(x)/x. The first two terms of that expansion are:

sinc(x) = 1 - (x^2)/6 + …

The arc length is 2rx and the chord length is 2rsin(x), where 2x is the angle subtended by the arc. Since the chord is one mile long and the arc is one foot longer, we have:

sinc(x) = 5280/5281

and the truncated series approximation above gives

x = sqrt(6/5281) = 0.03371 radians, or about 1.931 degrees.

Then the arc is part of a circle with perimeter

(5281/5280) x (180/1.931) = 93.23 miles,

so the radius r is 14.84 miles [this is a very flat arc!]

and the height h is 14.84 x (1 - cos(x)) miles = 44.49 ft.

Back in the day, engineers without benefit of cheap computers could have gotten this close using slide rules; i.e., four figures.

*Excellent post Richard.

You anticipated where I was going to take this.

*

Ether,

I’m not a programmer, but I love a good geometrical challenge.

The equations I came up with were: (r-h)sin(l/2r)=x/2 and 2rsin(l/2r)=(x^2/4+h^2)^1/2
where r was the radius of the curve, h was the height specified in your diagram, l was the length of the curve, and x was the length of the chord. To find the numerical solution, input values for l and x, and solve for h or r.

Was the “challenge” to write a program that found the solutions of these functions through guess and check? Or did I come up with a different solution? Plugging in the values on this thread with my trusty graphing calculator, I got a reasonable approximation of the answers posted on this thread.

Even getting a reasonable approimation is good enough, because there is no exact answer. I simplified until I ended up with a formula that looked something like this:

sin (theta/2) = (theta * Arc) / (2 * chord length)

With theta being the angle inside the two radii that would have gone from the center of the circle to the end of the cords.

That can only be solved iteratively, graphically, or by a lookup table (for those of us who remember not having calculators…)

Once you have theta, the height is trivial.

There is no explicit solution so you have to use iteration.

For folks like Christopher149 who know how to use modern math tools like Mathematica this is very simple to do, and you can get answers of arbitrary precision.

Mathematica is very expensive, but there are free CAS (Computer Algebra System) tools like Maxima which anybody can download and install for free. Maxima can also solve this problem with arbitrary precision (see my earlier post where I used Maxima to get 600 digits).

Attached are:

• The derivation of the equations

• A simple Delphi console app code which uses a binary search to find theta

If you replace sin(theta)/theta - 5280/5281 with a 12th degree Taylor expansion (see Richard’s earlier post about Taylor expansion), you can squeeze out 16 decimal digits of accuracy with this simple code using Intel’s X86 Extended Precision floats

*

code.pdf (25.7 KB)

code.pdf (25.7 KB)