Rope around the world


The magical extra foot

I want to take a moment and re-invent a centuries-old brain teaser, that immediately has both fun and practical results to those who deal with ropes, wires and distances.

Imagine 1 foot of rope lying on the ground, pulled taut and fitly anchored at 2 points a foot apart. That 1 foot of rope allows very little underneath it once anchored. Perhaps I could squeeze a small toy underneath, considering the slight stretch a rope may have. Next, I imagine replacing that 1 foot of rope with 2 feet of rope with ends that remain anchored a foot apart. Next I pick up the rope at the midpoint and pull it tautly upward, to see what I now may fit underneath. With a little thought, I see an equilateral triangle, 1 foot per side, and an ankle-biting barker, walzing underneath it. Nothing miraculous thus far. I added 1 foot, and it allowed me to fit larger objects, but still well under a foot tall, underneath that rope. If I write out the math, I have the following clearance.

\[ \frac{\sqrt{3}}{2} \approx 0.866 \text{ feet of clearance } \approx 10.4 \text{ inches} \]

Now, to make it interesting, I make anchors, 4 feet apart. Then, I attach the ends of a 5 foot rope to it, and see what height of objects this rope might permiss between the anchors. Curiously, I find exactly \( 1.5 \) feet clearance when I pick up the midpoint as before. How does 1 foot of slack buy me 1.5 feet of clearance?

Politely I dodge the hard question in favor of entertainment of experimentation. So instead I ask, how much more clearance could 1 foot of rope buy me?

I now (in hypothetical) install a pair of ground anchors exactly 12 feet apart, and grab 13 feet of rope. I raise the midpoint and what's this! I find I have 2.5 feet of clearance. This deserves a table.

Anchors
spacing
Slack
allowed
Top
clearance
0 + 1 = 12
4 + 1 = 1 12
12 + 1 = 2 12
24 + 1 = 3 12
40 + 1 = 4 12
60 + 1 = 5 12
\( x \) \( + 1 \) = 12 \( \sqrt{2x+1} \)
\( x \) \( + d \) = 12 \( \sqrt{2dx+d^2} \)
All units are in feet

Now, I wonder, What is the most large-and-ridiculous-yet-real-to-life idea I can imagine to test on this subject? I know, I will wrap a rope tautly around the world's equator, and then add 1 foot of slack. How much clearance may I now achieve? Would I be able to crawl or even walk underneath it? Could an elephant do likewise? A giraffe? Would a sauropod rearing in tripodal stance be able to grasp it? Could a redwood overcome the rope in its fall? Well such a question was posed by William Whiston, and perhaps a few others before him who may have also realized that our Earth was not flat.

How much rope are we talking about?

The circumference of the Earth, at the equator is about 24,901 miles. That's 131,477,280 feet in all! I need to patch together 131,477,280 + 1 feet of rope in my imagination to envision the question just posed. The corresponding radius of Earth to use in calculations is 20,925,259 feet.

Picturing the problem

This expanded imagination, gives rise to an entirely different puzzle, but certainly whose answer would be intuitively correlated. Upon understanding the line of thought which leads to this question, the magnitude of the answer must be astoundingly beyond belief. Something big must fit under this rope! I define a few quantities to explore the mathematics tersely.

\( R \)
The radius of the Earth \( \approx 20\,925\,259 \) feet
\( \delta \)
The slack added to the rope \( = 1 \) foot
\( \theta \)
Upon raising the rope to a taut configuration, the angle as measured from the center of the Earth, running from the peak point to the point it touches the Earth. I need to introduce this additional unknown, to create equations for my answer.
\( \epsilon \)
The answer, of how much clearance is created by adding \( \delta \) slack in the rope after wrapping the Earth tautly.

In my imaginary world where taut ropes may be pulled perfectly straight, I see the rope departing tangentially from the earth at some great distance, and then rising to the peak altitude \( \epsilon \). This gives me a right triangle with the following characteristics.

Leg 1
Measure from the center of Earth to its surface
Length \( = R \)
Leg 2
Measure along the rope running tangentially from Earth to the rope's peak.
Length \( = R \theta + \delta/2 \)
Hypotenuse
Measured from the center of the Earth to the peak altitude of the rope
Length \( = R+\epsilon \)

I can employ trigonometry and the Pythagorean theorem to get a few equations.

\[ R^2 + (R \theta + \delta/2)^2 = (R+\epsilon)^2\]
\[ \tan \theta = \frac{R \theta + \delta/2}{R} \]
\[ \sec \theta = \frac{R+\epsilon}{R} \]

Stepping through a tangle, slowly, in the right direction

Of the equations, I have 2 independent equations with 4 variables (2 inputs and 2 unknowns), so a solution exists for a given input. However, an exact solution in the traditional sense is elusive because I rely on a blackbox inverse function to remove a blackbox trigonometry function. But one of the unknowns, \( \theta \) stays hiding in one blackbox or the other despite attempts to cleverly rearrange variables.

With \( \theta \) being impossible to untangle, I can instead untangle all the other variables from \( \theta \).

\[ 2 \tan \theta - 2 \theta = \frac{\delta}{R} \]
\[ \sec \theta - 1 = \frac{\epsilon}{R} \]

With the first of these equations, I can iterate Newton's method to find the value that makes \( \delta = 1 \).

\[ f(x) = \tan x - x - \frac{\delta}{2R} \]
\[ f'(x) = \sec^2 x - 1 \]
\[ x_{i+1} = x_{i} - \frac{f(x_{i})}{f'(x_{i})} \tag{Newton's method} \]

The derivative is zero at \( \theta = 0 \), so I need to start elsewhere for Newton's method to be stable. I know \( \theta \in (0, \pi/2) \) so I need to pick a \( x_{0} \) in that range also. If I pick too low, Newton's method will bounce me out of the valid range. If I pick too high I will get excessive iterations before converging to an answer. So, now I look for an \( x \) such that \( \tan x - x \) equals the percent of the Earth's diameter by which I have lengthened the rope.

\[ \tan x - x \approx \frac{1}{41\,850\,518} \approx 2.389456684861105 \times 10^-8\]
\( x_{i} \) \( f(x_{i}) \) \( f'(x_{i}) \)
1.0000000 0.5574077 2.4255188
0.7701903 1.9984732e-1 0.9409731
0.5578066 6.6091569e-2 0.3892490
0.3880141 2.0721236e-2 0.1670646
0.2639829 6.3079206e-3 0.0730571
0.1776405 1.8924183e-3 0.0322321
0.1189283 5.6387164e-4 0.0142784
0.0794370

Newton's method seems to be taking forever, especially considering that I need to be within 1.2e-8 before my approximation of 1 foot is at least a number that rounds to 1 foot. Clearly, \( x_0=1 \) is way too high. I can try again blindly, or I can get much smarter by realizing a good first approximation.

Leap, and then look

I can approximate the functions of \( \theta \) by considering the Maclaurin series of some trigonometry functions.

\[ \tan \theta = \theta + \frac{\theta^3}{3} + \ldots \]
\[ \sec \theta = 1 + \frac{\theta^2}{2} + \ldots \]

I have just shown the terms up to \( \theta^3 \), because I am estimating the tangent function very near 0. I like the simple algebraic result that is achieved.

\[ \frac{\delta}{R} = 2 \tan \theta - 2 \theta \approx \frac{2 \theta^3}{3} \]
\[ \theta \approx \sqrt[3]{\frac{3 \delta}{2 R}} \]
\[ \theta \approx 4.154066764857954 \times 10^-3 \]

I return to Newton's method, to see how close I may approach an answer:

\( x_{i} \) \( f(x_{i}) \) \( f'(x_{i}) \)
4.154066764857954e-3 1.649335974028912e-13 1.725646920809069e-5
4.154057207074411e-3 3.794875331551159e-19 1.725638979901307e-5
4.154057207052420e-3 2.008989222880591e-30 1.725638979883035e-5
4.154057207052420e-3

Now, just a couple iterations firms up an astounding result. Having that 1 foot of slack offers over 180 feet of clearance.

\[ \epsilon = R (\sec \theta - 1) \approx 180.546 \text{ feet} \]

The amount of rope which leaves the ground to achieve this clearance is now also readily calculated — almost a thousand times more than the clearance itself.

\[ 2R \theta + \delta \approx 173\,850 \text{ feet} \]

Now, I can celebrate, but what if I want to figure the result for a yard of slack? A kilometer? A mile? Why not skip Newton's method and trigonometry altogether?

\[ \frac{\epsilon}{R} = \sec \theta - 1 \approx \frac{\theta^2}{2} \]
\[ \frac{\delta}{R} \approx \frac{2\theta^3}{3} \]

Therefore, a fast approximation is as follows.

\[ \epsilon \approx \frac{R}{2}\left(\frac{3\delta}{2R}\right)^{2/3} \]

I really love this algebraic approximation. In fact, if I stick to radians as my units (of both lengths and arcs), then I can drop \( R \) from the equations altogether. And how good is this approximation? That question is worthy of another table.

\( \delta \) \( \epsilon \approx \) \( \epsilon = \) (by Newton's)
1 foot 180.546 feet 180.546 feet
\( 2 \pi \) feet 614.763 feet 614.768 feet
1 mile 10.3684 miles 10.3765 miles
100 miles 223.379 miles 227.154 miles
Adding 1 radius of Earth
3,963 miles 2596.53 miles 3099.26 miles
Double the rope
24,901 miles 8841.4 miles 14280.5 miles

This table shows that adding rope up to 100 miles long, and my algebraic approximation is within 2 percent of the actual. In fact even adding a full radian is still within 20 percent of the actual. Also, the approximation is a lower bound, so it is in spirit of the original question, which seeks to confidently identify objects that do fit underneath the lengthened rope.