Difference between revisions of "Rope around the Earth"

Rope around the Earth
Field: Geometry
Image Created By: Harrison Tasoff

Rope around the Earth

This is a puzzle about by how much a rope tied taut around the equator must be lengthened so that there is a one foot gap at all points between the rope and the Earth if the rope is made to hover. Although finding the answer requires only basic geometry, even professional mathematicians find the answer strangely counter-intuitive. There is a related problem about stretching the rope taut again where the answer is even more surprising. A question similar to the first appeared in William Whiston's The Elements of Euclid circa 1702.

Basic Description

In this puzzle, we treat the Earth as though it were a prefect sphere, even though it actually bulges toward the equator. Suppose a rope was tied taut around the Earth's equator. It would have the same circumference as the Earth (24,901.55 miles). The question is: by how much would the rope have to be lengthened so that, if made to hover, it would be one foot off the ground at all points around the Earth?

Image 1[1]. The lengthened rope lifted by a point. Image not to scale.

Despite the enormous size of the Earth, and the 1 foot gap around the entire circumference, the rope would have to be lengthened by a mere 2π feet, or roughly 6.28 feet.

In fact, 2π feet is the answer regardless of the size of the ball around which the rope is wrapped.

Just as bizarre is what happens when one point on the lengthened rope is lifted up so that the rope is taut again, as in Image 1. The maximum clearance under the rope proves to be quite large. For the specific case of a rope looped around the Earth, a 2π foot extension would provide 614.771 feet of clearance if the rope were lifted. This is enough room to fit two Statues of Liberty under it, base and all. Unlike the previous question, however, this result is dependent on the size of the ball.

A More Mathematical Explanation

Note: understanding of this explanation requires: *High-school algebra and High-school geometry

The circumference of a circle is given by the equation: $C=2\pi\,\!r$, where r is [...]

The circumference of a circle is given by the equation: $C=2\pi\,\!r$, where r is the radius.
Image 2. Image not to scale.

In Image 2:

• Lrope 2 is the length of the extended rope.
• Cearth is the Circumference of the Earth and the original length of the rope (Lrope 1).
• Rrope 2 is the radius of the circle made by the extended rope.
• Rearth is the radius of the Earth and the original radius of the rope (Rrope 1).

When the rope is taut around the globe, its length equals the circumference of the Earth.

Eq. 1        $L_{rope 1}=C_{earth}=2\pi\,\!R_{earth}$

The puzzle states that we have to lengthen the rope and made the rope hover 1 foot of the surface of the earth. Lengthening the rope so that it is 1 foot off the ground at all points simply means changing the radius of the circle it forms from:  Rrope 1= Rearth to Rrope 2= Rearth+1 ft.

Thus:

Eq. 2        $L_{rope 2}=2\pi\,\!(R_{earth}+1)$

Distributing the 2 π yields:

Eq. 3        $L_{rope 2}=2\pi\,\!R_{earth}+2\pi\,\!=C_{earth}+2\pi\,\!$

The new length of the rope is merely 2 π feet longer than the original length. Indeed, one can see that the additional 2π is a result of distributing the 2π in Eq. 2, which yields an increase of 2π no matter what the radius of the ball is. Also, because the value of Rearth was not any number in particular throughout this proof, the answer didn't depend on the specific radius of the Earth in any way. Hence, this 2π extension would be the same for a ball, planet, or star of any size.

It should be noted that the Earth is not a perfect sphere. Not only does it bulge out around the equator, but mountains and valleys give it a rough surface. This roughness affects the amount of slack needed, as might be expected. Follow this link to an discussion of this in the Why It's Interesting section.

Maximum Height of Rope

Note: A familiarity with trigonometry, series, and approximations is recommended for this section.

Were the lengthened rope again to be held taut by raising it at an arbitrary point (as shown in Image 3), what would be the distance from this point to the surface of the earth?

Image 3[2]. Image not to scale.

In Image 3:

• X1 is the distance between where the rope leaves the Earth's surface and the highest point on the taut rope.
• Xo is the ground distance along the curved surface of the Earth from the point where the rope leaves the globe, to the point below the apex of the rope.
• R is the radius of the globe.
• h is the height of the apex of the rope above the ground.
• θ is the angle formed between the line from the center of the Earth to the apex and the line from the center of the Earth to the point where the rope leaves the ground.

In the puzzle the slack added to the rope is 2π feet. Now let l represent the this slack because, in a different version of the puzzle, l can be different. Now we will develop a formula for l, the amount of slack, in terms of h, the maximum clearance the slack can provide. We start with h because it appears clearly in the diagram, where l is less visible. We hope later to invert this in-order to solve for h in terms of l. First, we will do this for the case where l = 2 π.

Notice that triangle ABD is a right triangle, with (R + h) as its hypotenuse. Thus, using , we know that:

$(R+h)^2=R^2+X_1^2$

This is equivalent to:

Eq. 4         $X_1=\sqrt{(R+h)^2-R^2}$

To find the length of Xo, we must remember what the length of an arc (Larc) is:

Eq. 5         $L_{arc}=r\theta\,\!$

where r is the radius of the circle, and θ is the angle, in radians, formed between two radii from the center of the circle to the endpoints of the arc. For more on Arcs, see Arcs.

In Image 3, Xo is the arc whose length we want to find; hence, it can be substituted in for Larc in Eq. 5. Also, since cos-1$\ \left (\tfrac{R}{R+h} \right )$ represents θ, the angle whose cosine is $\tfrac{R}{R+h}$, we can replace θ with cos-1$\ \left (\tfrac{R}{R+h} \right )$. For more information on cos-1 see Inverse Trig Functions.

Thus, Eq. 5 is equivalent to:

Eq. 6         $X_o=R\,\cos^{-1} \left (\frac{R}{R+h} \right )$.

Now that we have a formula for Xo, were are going to relate Xo to X1 . Because the rope is taut around most of the earth, all the slack goes to lengthening arc AC and turning it into segments AB and BC. Since we know that we lengthened the rope by 2π feet, we know that 2X1= 2Xo + 2π, because 2X1 - 2Xo is the extra slack put in to the rope. Thus, more generally:

Eq. 7         $2X_1=2X_o+ \mathit{l}$

Solving for l in Eq. 7 yields:

Eq. 8         $\mathit{l}=2 \left (X_1-X_o \right )$

Substituting Eq. 4 and Eq. 6 into Eq. 8 yields:

Eq. 9         $\mathit{l}=2 \left ( \sqrt {(R+h)^2-R^2} -R\,\cos^{-1} \left (\frac{R}{R+h} \right ) \right )$

Because h is both in the of the cos-1 and elsewhere in the problem, we can't isolate it and solve for it explicitly. Nevertheless, running Eq. 9 through a with l = 2 π and R = R earth = 20925524.9 feet one finds h ≈ 614.771 ft as the height that a 2π foot extension yields.

Eq. 9 can, however, be used to find the amount of slack needed to produce a certain height. For instance, if we set h = 35,000 feet (the average cruising altitude for commercial aircraft), Eq. 9 tells us that l = 2697.06 feet. Thus, Roughly 3,000 feet of slack must be added to the rope to that, when lifted, a plane could fly under it on its normal flight path.

If the puzzle were different, having a different amount of slack added to the rope, we could still find h if given a value for l. We can create a general formula that will find h for any value of l, in addition to when l = 2π. As mentioned above, due to the cos-1 (R / (R + h)) there is no explicit formula to find the height. Nevertheless, using series, on can obtain very good approximations. Using series approximations one can derive the following formulas for height achieved by lengthening the rope by length l; the first is a 1st order approximation, and the second, a more accurate 2nd order approximation. The approximations are more accurate the smaller the added slack is compared to the original radius of the circle.

Image 4. The series approximations for a large value of R relative to l. Image created by Harrison Tasoff

Eq. 10         $h \left (\mathit{l} \right ) \approx \frac{1}{2} \left ( \frac{3}{2} \right )^{\left ( \frac{2}{3} \right )}R^{ \left ( \frac{1}{3} \right )} \mathit{l}^{ \left ( \frac{2}{3} \right )}$

Eq. 11         $h \left (\mathit{l} \right ) \approx \frac{1}{2} \left ( \frac{3}{2} \right )^{\left ( \frac{2}{3} \right )}R^{ \left ( \frac{1}{3} \right )} \mathit{l}^{ \left ( \frac{2}{3} \right )} + \cfrac {9 \left ( \frac{3}{2} \right )^{\left ( \frac{1}{3} \right )}}{80 R^{ \left (\frac{1}{3} \right )}}\ \mathit{l}^{\left ( \frac{4}{3} \right )}$

Where l is the added slack, h(l) is the height as a function of l, and R is the initial radius of the rope circle.

Taking l = 2π feet of slack and R = R earth = 20925524.9 feet from the initial problem the first approximation yields 614.766, demonstrating the accuracy of these approximations. The second approximation yields 614.771, which is even more more accurate than the first.

Image 4 illustrates these approximations, with l on the x-axis and h(l) on the y-axis. Eq. 10 appears in red and Eq. 11 in blue. In this example, the radius R of the ball is an arbitrary number that is very large relative to l. Notice that for small values of l, Eq. 10 and Eq. 11 are indistinguishable. It is only once l starts to become a decent percentage of R that Eq. 11 becomes noticeably more accurate than Eq. 10. When R= 5000, as it does in this graph, Eq. 11 becomes appreciably more accurate than Eq. 10 when l ≈ 150, or roughly 3% of R. At this point, Eq. 10 outputs 30827.651, while the more accurate Eq. 11 outputs 30833.653, a difference of 6.002.

Here is another way to illustrate the varying degrees of accuracy in these approximations. The most accurate value the root finder yields is 614.7709915007723. Comparing this to the value found by the 1st order approximation (614.7655785603816), we find that the latter is accurate to five significant digits when rounding is accounted for. The 2nd order approximation (614.7709968720138) is accurate to seven significant figures when rounding is accounted for. Hence, though in absolute terms, both approximations are quite accurate, the 2nd order approximation is 100 times more accurate relative to the 1st order approximation.

Also, as one can see the approximations climb quickly from l = 0. This steep ascent means that even small values of l output large values for h(l), or, even small amounts of extra slack yield large increases in maximum height.

On the scale of the Earth, there is a decent margin of error in measurements: the earth is not a perfect sphere, nor perfectly smooth, ropes stretch with strain, objects expand and contract with heat. The result is that these approximations are far more accurate and precise than our measurements can ever be.

Why It's Interesting

At first this puzzle is interesting because of its non-intuitive results. It becomes even more interesting if we can then make them intuitive after all, and indeed and we can. Though it may seem that 2π, or roughly 6.28, feet is a minuscule amount of extra rope needed to to produce such a considerable result, a look at the ratios will show otherwise.

The radius of the Earth is roughly 20,920,000 feet, though this varies because the Earth is not a perfect sphere. There is 1 foot of difference between the radius of the circle made by the lengthened rope and the radius of the Earth. This foot of difference is a mere fraction of the radius of the Earth: about five one-hundred millionths, or .000000047, of the Earth's radius. A foot doesn't seem so large anymore.

Similarly, 2 π feet is .000000047 of the circumference of the Earth (which is about 131,000,000 feet). And, unsurprisingly, the ratio of 1 foot to the Earth's radius is the same as that of 2 π feet to the Earth's circumference.

So, in this perspective, a small change in the length of the rope yields a proportionally equivalent small change in the radius of the rope circle.

Image 5. A convex polygon and a concave polygon of the same type.

Another surprising aspect of this puzzle is that, it is not the presence of mountains that affects the amount of extra slack needed, but rather, their shape. The steep mountains and valleys cause the earth to have a concave cross-section, rather than a convex one, when considered in detail. See Image 5 for an illustration of the difference between convex and concave. If the Earth had a convex cross-section, then the amount of slack needed to raise the rope one foot above the ground would still be 2π feet, regardless of how many sides it has in cross-section. In fact, the equator could even be a square and the 2π feet of slack needed still holds true.

Image 6. A square with a blue rope 1 unit away from it at all points. Notice that the four quarter circles together require adding 2π units of slack to the rope's length.

Image 6 shows a blue rope around a square of side length 1. The rope is 1 unit away from the square at all points.

Notice that the straight portion of each side of the larger figure is 1 unit long. Each "corner" of the larger figure is a quarter circle with a radius of 1 unit. The length of each quarter circular segment is 2π(1)/4, one quarter that of a full circle of radius 1. Thus the perimeter of the larger figure, Which is the length of the blue rope, is:

$\text{Length} = 1+1+1+1 + \frac{2\pi(1)}{4} + \frac{2\pi(1)}{4} + \frac{2\pi(1)}{4} + \frac{2\pi(1)}{4} = 4 + 2\pi$

The perimeter of the small square is 4.

Subtracting the perimeter of the smaller figure from that of the larger, you find a difference of (4 + 2π) - 4 = 2π units of slack.

We will prove that this is the case for any convex shape.

We start with the equation for the measure of each angle of a . Because all angles are the same measurement, the polygon is automatically convex. Notice in Image 5 that the concave shape does not, and indeed cannot, have all of its angles be of equal measurement. n is the number of sides the polygon has, and θ is the measurement of an .

Eq. 12         $\theta= \frac{\left (n-2 \right ) 180^o}{n}$

The derivation of this is based on triangles, and can be seen here.

Image 333. A regular polygon, in this case a hexagon, with two equal length line segments perpendicular to two adjacent sides located at the corner between them. θ is labeled with the equation that defines it, and the two squares in the right angles identify them as 90°.

To remain 1 unit away from the shape at all points, the rope will have to curve around each corner in a circular path of radius 1. This path starts as soon as the rope reaches the end of a side, and lasts until it reaches the beginning of the next side, as in Image B with the square. Finding the measure of φ in Image C will tell us how much of a circle is required to go around each corner.

All of the angles in the upper left-hand corner of Image C total to 360° (one full revolution). Subtracting the other angles from 360° will yield the measurement of φ.

Eq. 13         $\phi = 360^o - 90^o - 90^o - \theta$

Which simplifies to:

Eq. 14         $\phi = 180^o - \theta$

Thus, φ is the to θ. Because Eq. 13 and Eq. 14 make no mention of the other interior angles of the shape, this result is true for each corner of a non-regular shape as well.

We now can calculate the measure of φ. Because a circle has 360°, as it is one full revolution, each corner of a figure requires a section that is $\tfrac{\phi}{360^o}$ of the entire circle. We now seek to find how much length these segments add when all of them are put together.

For any convex shape with n sides, the measurements of all of the interior angles will add to the same value as those of any other n sided shape as long as it is convex. For instance, the interior angles of every triangle always add to 180°, while those of all add to 360°, and so on. As a result, the following work will apply equally well to non-regular shapes as it does to regular ones.

A shape with n sides will have n corners, so there will be that same number of these circular segments as there are sides to the shape. The total angle measurement of all of the φs is:

Eq. 15         $\phi_{total}=n \phi$

Substituting in what φ equals from Eq. 14:

Eq. 16         $\phi_{total}=n (180^o - \theta)$

We can substitute in the value we found for θ in Eq. 12:

Eq. 17         $\phi_{total}=n \left (180^o - \frac{ \left ( n-2 \right ) 180^o}{n} \right )$

Distributing the n, we get:

Eq. 18         $\phi_{total}=180^on - \frac{n \left (n-2 \right ) 180^o}{n}$

One n' in the cancels out with the n in the , and then we distribute the 180° and minus sign:

Eq. 19         $\phi_{total}=180^on - 180^on + 2 \left (180^o\right )$

As you may see, the 180°n will cancel with the -180°n. Once we distribute the 2, we have:

Eq. 20         $\phi_{total}=360^o$

Indeed, the total measurement of all of the φs is exactly 360°, so all the circular segments add to exactly one whole circle. Accordingly, the total amount of extra rope needed for all of the circular corner sections is the circumference of this circle with a radius of 1 unit. Hence, 2π units of slack. We have now proven that for any convex polygon (regular and irregular), you can increase a the distance of a rope from of a polygon by 1 unit by adding 2π units of slack to the rope.

References

• Pickover, C. A. (2009). The Math Book. New York: Sterling Publishing Co.
• (2009, March 3). Roping the Earth. Message posted to: