Rope around the Earth: Torsten Sillke
Problem A: [Langman 1935]
Suppose the earth is a sphere of radius r. A flexible belt is constructed
around the equator, but large enough to encircle a sphere with radius one
inch greater. If a vertical pole is placed under the belt at one point,
drawing it taut, how high must it be, and how far from this pole does
the belt first touch the earth? Assume: r = 4000 miles.
Problem B: [Hess 1997, 1999]
Imagine a rubber band streched around the world and over a building.
Given that the width of the building is 125 feet and the rubber band
has to strech an extra 10cm to accommodate the building, how tall is
the building? (Use 20902851 ft for the radius of the earth.)
Problem C: [Norris 1901]
Two cities are 200 miles apart. To what height must a man ascend from
one city in order that he may see the other, supposing that the
circumference of the earth is 25000 miles? Find the solution by using
geometry.
Problem D1: [Honsberger 1981]
Imagine a flat strech of straight railway track AB, 5000 feet long,
which is immovably fastened down at each end. In the heat of the summer
the track expands 2 feet, causing it to buckle. Assume it bends
symmetrically, how high do you think the bulge rises above the ground
at the middle?
Problem D2: [OMG]
A section of railway track 5000 meters long was laid in the desert.
Because of the heat during the day, the workmen put the track down
during the cool of the night and securely fastened each end. In the
heat of the following day the section of track expanded by 1 meter
in length. If the track bowed upwards, how high would the center of
the track be above the ground level?
Problem D3: [Hal91]
You have a flexible piece of railroad track which is 1 mile + 1 foot
long. You have a spot for it which is 1 mile long, so you jam it
into place, fixing the ends of the track 1 mile apart. Since the
track is too long, it bulges up in the middle; assume that it bends
into an arc of a circle. How far off the ground is the middle of
the track? [Halmos, problem 2 I, page 19]
Solution B: [Hess]
Let w = width of the building; h = height of the building;
r = the earths radius; and d = stretch in the band due to
the building. Then
2 tan(beta) = w / (r + h)
r tan(alpha) = sqrt(2rh + h^2 + w^2/4)
d = 2r tan(alpha) + w - 2r (alpha + beta).
These equations can be iterated to produce h = 85.763515 ft.
Solution D:
Let t = length of the track; Delta = increment of the length.
Let d = Delta / ( t + Delta ) then we have to solve
d = 1 - sin(x) / x
+--__ x
a ''--__
+----s-----+
| / a = 1 - cos(x)
| / s = sin(x)
| / c = cos(x)
| /
c /1
| /
| /
| /
| /
X
x = w (1 + O(d)) with w = sqrt(6 d).
a = 1/2 w^2 ( 1 + O(d) )
Now the radius of the arc is R = ( t + Delta ) / (2x).
height = R * a = 1/4 ( t + Delta ) sqrt(6d) ( 1 + O(d) )
= 1/4 sqrt(6 t Delta) ( 1 + O(d) ).
References:
- P. Abbott;
In and Out: Acton's Railroad Problem,
Mathematica Journal 7 (2000) 448-450
- F. S. Acton;
Numerical Methods That Work,
2nd printing,
Mathematical Association of America 1990
(Railroad Track Problem)
- Harry Langman;
Problem E 103: (rope around the earth)
American Mathematical Monthly 41 (1934) 390 problem by Langman
American Mathematical Monthly 42 (Jan 1935) 46 solution by E. C. Kennedy
- Paul R. Halmos;
Problems for Mathematicians, Young and Old,
MAA, Dolciani Mathematical Expositions 12, 1991
ISBN 0-88385-320-5
[problem 2 I, page 19, Railroad Track Problem]
- Heinrich Hemme;
Der Wettlauf mit der Schildkr\"ote,
G\"ottingen, 2002, Vandenhoeck & Ruprecht, ISBN 3-525-40740-8
problem 86: Das Seil um den \"Aquator
- Richard I. Hess;
Puzzles From Around the World, April 1997, self-distributed.
problem 33: A rubber band around the world
- Richard I. Hess;
Puzzles From Around the World, in
Elwyn Berlekamp and Tom Rodgers;
The Mathemagician and Pied Puzzler,
1999, p53-84
problem H4: A rubber band around the world
- Nick Hobson;
Equatorial belt,
Nick's Mathematical Puzzles 4,
http://www.qbyte.org/puzzles/p004s.html solution
- Dr. Math;
Segments of Circles: The Arc, Chord, Radius, Height, Angle, Apothem, and Area
http://mathforum.org/dr.math/faq/faq.circle.segment.html
Case 1: Given Arc and Chord.
- S. F. Norris;
Problem GE 166:
American Mathematical Monthly 8 (1901) 127 problem by S. F. Norris
American Mathematical Monthly 9 (1902) 75 solution by D. Northrop
- Marie Berrondo;
Les jeux mathematiques d'Eureka, Bordas, Paris 1979.
(German: Eureka's mathematische Spiele, Hugendubel 1986;
Fallgruben F\"ur Kopf-F\"ussler, Fischer-Logo 8703, 1989;
Problem 4.22 Der Leuchttum - wie weit ist der Horizont)
- Ross Honsberger;
Games, Graphs, and Galleries,
Problem: The Bulging Railway (Murry Klamkin)
In: The Mathematical Gardner, ed. by David Klarner;
Prindle, Weber & Schmidt/Wadsworth, 1981. Pp. 274
In: Mathematical Recreations - A Collection in Honor of Martin Gardner,
Dover Publ. 1998, p274.
- OMG;
Problem 17.2.2 (railway track problem)
Ontario Mathematical Gazette 17 (1978/2) 58 problem
- Eric Weisstein;
World of Mathematics,
Railroad Track Problem,
http://mathworld.wolfram.com/RailroadTrackProblem.html
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Solving the Equation
tan(x) - x = w^3/3
Taylor series of tan is
tan(x) = x + 1/3 x^3 + 2/15 x^5 + 17/315 x^7 + O(x^9)
So we have the approximate equation
1/3 x^3 + O(x^5) = 1/3 w^3
=> x^3 ( 1 + O(x^2) ) = w^3
=> x ( 1 + O(x^2) ) = w
=> x = w ( 1 + O(x^2) ) = w ( 1 + O(w^2) ) = w + O(w^3).
Using more terms of the series of tan we get the relation
2 2 29 4 14 6 8
x = w ( 1 - -- x - ---- x - ---- x + O(x ) )
15 1575 3375
____________
3 / x^3 / 3
= w \ / ------------
\/ tan(x) - x
Starting with x0 = w we can plug in this at the right hand side
and get an better approximation for x. At each step we get a
further term of the series.
x0 = w + O(w^3)
x1 = w - 2/15 w^3 + O(w^5)
x2 = w - 2/15 w^3 + 3/175 w^5 + O(w^7)
. . .
2 3 3 5 2 7 16 9 11
x = w - -- w + --- w - ---- w - ------ w + O(w )
15 175 1575 202125
Another way to increase the accuracy of a approximate solution
is using Newton's method. For f(x) = tan(x) - x - w^3/3 we want
to solve f(x) = 0. With the derivative f'(x) = tan^2(x) we iterate
x(n+1) = x(n) - f(x(n))/f'(x(n)).
Again we start with x(0) = w.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Solving the Equation
1 - sin(x)/x = w^2/6
Taylor series of sin is
sin(x) = x - 1/6 x^3 + 1/120 x^5 - 1/5040 x^7 + O(x^9)
So we have the approximate equation
1/6 x^2 + O(x^4) = 1/6 w^2
=> x^2 ( 1 + O(x^2) ) = w^2
=> x ( 1 + O(x^2) ) = w
=> x = w ( 1 + O(x^2) ) = w ( 1 + O(w^2) ) = w + O(w^3).
Using more terms of the series of sin we get the relation
1 2 23 4 13 6 8
x = w ( 1 + -- x + ----- x + ------- x + O(x ) )
40 67200 4838400
____________
/ x^3 / 6
= w \ / ------------
\/ x - sin(x)
Starting with x0 = w we can plug in this at the right hand side
and get an better approximation for x. At each step we get a
further term of the series.
x0 = w + O(w^3)
x1 = w + 1/40 w^3 + O(w^5)
x2 = w + 1/40 w^3 + 107/67200 w^5 + O(w^7)
. . .
1 3 107 5 3197 7 9
x = w + -- w + ----- w + -------- w + O(w )
40 67200 24192000
Therefore the equation 1 - sin(x)/x = d has the solution
_____ 3 321 2 3197 3 445617 4 1766784699 5 6
x = \/ 6 d ( 1 + -- d + ---- d + ------ d + -------- d + ------------ d + O(d ) )
20 5600 112000 27596800 179379200000
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Inverting a Taylor series is called Lagrange Inversion.
Comtet [Com1974] gives an explicit formular to invert a series.
Theorem 3.8.F. Let a be an integer >=1. For
s = t ( 1 - Sum_{m>=1} x_m t^(a m)/m! ), we have
t = s ( 1 + Sum_{m>=1} y_m s^(a m)/m! ), where
y_m = Sum_{1<=k<=m} (a m + k)_(k-1) B_{m,k}(x1, x2, ...)
(x)_k = x*(x-1)* ... *(x-k+1) the falling factorials and
B_{m,k}(x1, x2, ...) are partial exponential Bell polynomials.
Examples: B_{1,1} = x1, B_{2,1} = x2, B_{2,2} = x1^2.
This Theorem shows that the first term is always easy to find.
Com74: Louis Comtet;
Advanced Combinatorics,
Reidel Publishing Company, Dordrecht (1974)
Chap. 3: Identities and Expansions
Chap. 3.8: Inversion Formula of Lagrange
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Measures of the Earth:
1. Ellipsoid after Friedrich Bessel 1841:
major semi axes a = 6,377,397.1550 m
minor semi axes b = 6,356,078.9630 m
sphere of equal volume: 6,370,283.1583 m
sphere of equal area: 6,370,289.5102 m
sphere of equal dist(pol): 6,366,742.5203 m
1. Ellipsoid after Hayford 1924 (International 1924):
major semi axes a = 6,378,388.0000 m
minor semi axes b = 6,356,911.9461 m
sphere of equal volume: 6,371,221.2659 m
sphere of equal area: 6,371,227.7113 m
sphere of equal dist(pol): 6,367,654.5001 m
1. Ellipsoid after Krassovsky 1940/1948:
major semi axes a = 6,378,245.0000 m
minor semi axes b = 6,356,863.0188 m
sphere of equal volume: 6,371,109.6937 m
sphere of equal area: 6,371,116.0829 m
sphere of equal dist(pol): 6,367,558.4969 m
1. Ellipsoid World Geodetic System of 1972:
major semi axes a = 6,378,135.0000 m
minor semi axes b = 6,356,750.5200 m
sphere of equal volume: 6,370,998.8588 m
sphere of equal area: 6,371,005.2495 m
sphere of equal dist(pol): 6,367,447.2486 m
1. Ellipsoid World Geodetic System of 1984:
major semi axes a = 6,378,137.0000 m
minor semi axes b = 6,356,752.3142 m
sphere of equal volume: 6,356,752.3142 m
sphere of equal area: 6,371,007.1809 m
sphere of equal dist(pol): 6,367,449.1458 m
2. Distance North pole to equator = 10000 km (First definition for the meter)
Circumference = 40000 km.
Radius = 6366.198 km
3. Circumference = 60*360 Int. Sea Miles
Radius = 3437.747 Sea Miles = 6366.707 km
(1 international sea mile = 1.852 km)
4. Circumference = 25000 Miles
Radius = 3978.874 Miles = 6403.376 km
(1 statute mile = 1760 yard = 1760*36 inch = 1609.344 meter)
5. Circumference
Radius = 4000 Miles = 6437.376 km
(1 statute mile = 1760 yard = 1760*36 inch = 1609.344 meter)
References:
- Mentor Software Freebies,
Ellipsoid Radii, Version 1.0, 1999
www.mentorsoftwareinc.com
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
1) Assume the equator is a perfect sphere, with circumference 40.000
kilometers. Suppose a rope (not stretchy) is tied tightly around it. Now if
one meter of rope is added, and all around the equator the rope is lifted to
the same height, what height above the earth will the rope be?
2) The same conditions as the first problem, but now the rope is pulled at
one point as high as possible above the earth. What height will be reached?
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
|> 2) The same conditions as the first problem, but now the rope is pulled
at one
|> point as high as possible above the earth. What height will be reached?
I don't know if this is elegant, but here it is:
Suppose the rope is pulled up to a height h above the earth at P. On each
side of P the rope lies on a straight line tangent to the earth's surface.
Let Q be one of the points of tangency, and C the centre of the earth.
Then CQP is a right triangle, with PC = r+h and QC=r where r is the radius
of the earth. Let t be the angle QCP. Then PQ = r tan t = r t + 1/2
(since this side of the triangle has half of the extra metre of length)
= sqrt((r+h)^2-r^2) = sqrt(2 r h + h^2). Since h is much smaller than r,
we can approximate this as sqrt(2 r h) (to leading order in r). In all
that follows, "=" is to be understood as "approximately" (ignoring higher
order terms in r). Thus h = r/2 tan^2 t where tan t = t + 1/(2 r).
Now tan t = t + t^3/3 + ... so t^3/3 = 1/(2 r), i.e. t = (3/(2 r))^(1/3) and
h = r/2 t^2 = r^(1/3) 3^(2/3) / 2^(5/3) = about 121.43 metres.
Robert Israel israel@math.ubc.ca
Department of Mathematics (604) 822-3629
University of British Columbia fax 822-6074
Vancouver, BC, Canada V6T 1Y4
The answer to the first problem does not depend on the size of the earth. It
is
1/(2 pi) =0.15915... m, or about 16 cm.
The answer to the second problem does depend on the size of the earth. I
will
use a better value for c, the circumference of the earth. The radius of the
earth at equator is
r = 6378160 m
c= 2 pi r = 40075161 m
With a little thought we see that
r tan(x) - r x = 1/2
where x is half the angle ,in radians, between the two points on the equator
where the rope leaves the surface of the earth. The units of the equation
are
meters.
or
tan(x) -x = 1 / (2 r) = pi / c
Since tan(x) = x + x^3 / 3 + 2 x^5 / 15 + ...
we can get an approximate value a for x from
a^3 / 3 = pi / c
a = (3 pi / c)^1/3 = 0.006172559
We can iterate the equation
x = atan(x + pi / c)
and it converges slowly to x = 0.00617252811726
The maximum height h of the rope above the earth is gotten from
(r + h) cos(x) = r
h = (r / cos(x) - r) = 121.5062 or about 121.5 meters.
The answer to problem 2 is about 763 times larger than the answer to problem
1.
Harry
--
| Harry J. Smith
| 19628 Via Monte Dr., Saratoga, CA 95070-4522, USA
| Home Phone: 408 741-0406, Work Phone: 408 235-5088 (Voice Mail)
| E-mail: HJSmith@ix.netcom.com on the Internet via Netcom NetCruiser
--
On Sun, 30 Apr 2000 18:41:33 -0700, in sci.math you wrote:
>For the problem which Peter gave, where s = 1 meter and
>r = 4*10^7/(2*pi) meters, my formula gives a height of about
>121.43 meters. I think this is quite a good approximation,
*Excellent*, in fact, and to be expected. I can't quote earlier posts
not vailable to me now. However, here's a full solution for what it's
worth. No approximation except for what Maple does naturally:
Given Earth Circumference = 40000.
A string encircles the earth but 1m is added.
If the string is pulled taught, how high is it from the
surface of the earth?
Diagram:
A circle radius R, height h, tangent x, and angle in a right triangle
so formed at the center of the circle, A.
x^2 = (h+R)2 - R2 = h2+ 2hR
Large arc length = 2 pi R - 2 R A
So, 2 pi R - 2 R A + 2x = 2 pi R + .001
and x = R A + .0005
or x/R = A + .0005/R
That is:
arctan(x/R) + (.0005/R) = x/R
Now, R = 40000/(2pi) = 20000/pi
So, arctan(x* pi/20000) = (x* pi/20000) - pi*.0005/20000
Maple V does this:
fsolve(arctan(x*Pi/20000) - Pi*(x-.0005)/20000 = 0,x);
x = 39.3206313
fsolve(h^2 + 40000/Pi*h - 39.3206313^2=0,h);
h = .1214301981, or 121.43m
Dan.
-------------------
O.K. The solution IS straight-forward, though it is non-algebraic. I don't
think there is a closed form solution.
It's very easy to show the relationships involved with a simple diagram,
but I will try to explain things in text.
The easiest way to approach the problem is to work it backwards and then
use some iterative method to bring it forwards again. The resulting
function is very well behaved and so just a few iterations will yield a
solid approximation.
By working it backwards, I mean that we assume we have values for R and h
and that we want to know the value of d. R is the radius of the circle, d
is the amount by which the radius of the rope exceeds R and h is the
distance from the apex of the resulting teardrop to the circle.
Draw a circle of radius R centered at the origin.
From a point X on the positive x-axis draw a tangent line back to the upper
half of the circle. From this tangency point (which we'll call T and which
will always be in the first quadrant) draw a line to the origin. The angle
of elevation of this line from the positive x-axis is theta. There is also
a mirrored version of the tangent line from a point in the fourth quadrant
to X.
The length of line segment TX we'll call L.
The circumference of the circle is clearly
C=(2pi)R
and the length of the rope is clearly
D=(2pi)(R+d)
D=(2pi)R+(2pi)d
The length of the path around the 'raindrop' is simply
D=2L+2R(pi-theta)
D=2L+(2pi)R - (2theta)R
Equating the two equations for D we find
(2pi)R+(2pi)d=2L+(2pi)R - (2theta)R
(pi)d = L - (theta)R
d = (L - (theta)R)/(pi)
d = (R/pi)(L/R - theta)
By the Pythagorean Theorem, the length of TX is
L = sqrt(X^2-R^2)
L = sqrt((R+h)^2-R^2)
L = sqrt(h^2 + 2hR)
The tangent of theta is then
tan(theta) = L/R
= sqrt(h^2 + 2hR)/R
= sqrt( (h/R)^2 + 2(h/R) )
Let's define 'a' to be
a = tan(theta) = sqrt( (h/R)^2 + 2(h/R) )
So now we have
d = (R/pi)( a - atan(a) )
Summarizing what we've got so far:
d(h) = d(a(h))
a(h) = sqrt( (h/R)^2 + 2(h/R) )
d(a) = (R/pi)( a - atan(a) )
This is a closed form solution for the problem worked backwards. To go
forwards we simply use some numerical method such as Newton's Method or
even Linear Interpolation to find the value of h that corresponds to a
given value of d.
To put bounds on the possible values of d, as well as to provide an initial
guess if desired, we can look at two limiting cases:
What would h be if the rope were kept in contact with the entire circle and
the remainder brought out along the x-axis? The actual value of h must be
greater than this.
What would h be if the rope were cut and each end extended horizontally
from the top and bottom of the circle? The actual value of h must be less
than this.
These two cases directly yield:
h = (pi)d + alpha
where
0 < alpha < R(pi/2 - 1)
Using the simplest Linear Interpolation method I could think of, the
solution converges very rapidly for d/R ratios greater than about 0.01. 1%
in 5 iterations and 1ppm in 13. The greater the d/R ratio, the faster the
convergence. For d/R ratios greater than a little more than 0.1, the above
limits describe a range of about 1%.
Iterations to find the solution to an error of less than:
d/R 1% 1 ppm
1E-6 144 392
1E-3 12 34
0.01 4 13
0.1 2 4
1 0 2
>10 0 1
There are a number of ways of looking at the results to gain insight on the
behavior. You will see that for d/R greater than about 10, h=d(pi) becomes
a pretty good approximation.
I don't know what the subtleties were that you spoke of - other than that
the problem I am talking about is for a rope that doesn't stretch (and is
infinitely thin). As d/R gets small, any real rope (supporting any load at
all) is going to be placed under a great deal of tension and will stretch
quite a bit - but that is a different problem.
William L. Bahn (bahn@pcisys.net)
alt.math.recreational
1998/01/22
----
To make life simpler, I've relabeled your angle theta as angle B.
Virtually every relevant aspect of this problem can be described in terms
of this angle, so it'll be used liberally and the shortened name simplifies
life a bit. O represents the origin.
The point T on the circle has coordinates (R*cosB, R*sinB), the segment OT
has slope tanB, and the tangent line is perpendicular to this and hence has
slope -(1/tanB). With this one can determine an equation for this tangent
line and find that it has an x-intercept of (R*secB,0). This tells us
point X has coordinates (R*secB, 0).
At this point, we see that if we knew angle B, the height H that the
strings lies above (to the right of, actually) "the earth" is R*secB-R, or
H=R(secB-1).
An equation determining angle B can be found exactly as you did by equating
the circumference of the circle of string which had radius R+d to the
length of the string in the "teardrop" shape in our picture. To do this,
we need length L between points T and X. Since T=(R*cosB,R*sinB) and
X=(r*secB,0), the distance is easily computed using the distance formula.
Doing so and simplifying the result gives length L as R*tanB.
So, equating the circumference of the original circle: 2*Pi*(R+d)
to the length of the teardrop: 2*L+2*Pi*R*(Pi-B)
Substituting L=R*tanB and simplifying gives:
tanB-B=(Pi*d)/R.
Unfortunately, this equation can not readily be solved for angle B, but
Newton's Method (or your favorite "root finding algorithm") can be used to
approximate a solution. A little caution has to be used selecting an
initial value due to the nature of the function tanB-B.
So, ultimately we have H=R*(secB-1) where B is the solution to
tanB-B=(Pi*d)/R.
Dean