Yesterday I recalled an interesting brainteaser that I saw in a book a few years ago. It went something like this: Suppose a rope was tightly wrapped around the equator (picture A).

Then suppose that the length of the rope was increased by only a single yard. A mouse would be able to fit between the rope and the earth not just at one point (picture B), but at any given point around the earth (picture C). By working out the math, I could prove that the radius of the loop of rope increases by 5.73 (18/pi) inches. Doesn't this seem contrary to what one would expect?

 

Doesn't this seem contrary to what one would expect?

If you proved it mathematically, how can it be contrary to what you expect?

John

 

If you proved it mathematically, how can it be contrary to what you expect?

John

It makes perfect sense mathematically, but to me (and I think many others) it seems odd. I trust the math, but I have a hard time understanding "where did all that rope come from?" Scientific discoveries can be shocking.

 

To say it another way, I find it difficult to wrap my mind around this because the earth is so massive. How can adding a single yard of rope produce a ring with an area larger than 2 square miles?

 

Don't believe it! Show us the maths.

 

Don't believe it! Show us the maths.

OK, here's my first attempt at using LaTeX (all dimensions are in yards; c stands for circumference, d for diameter, r for radius):

\(c_{earth} = \pi \times d_{earth}\\
d_{earth} = 2 \times 3959 \times 1760\\\\
c_{earth} + 1 = \pi \times d_{ring}\\
d_{ring} = \frac{c_{earth}+1}{\pi} = \frac{c_{earth}}{\pi} + \frac{1}{\pi}\\
r_{ring} = \frac{d_{ring}}{2} = \frac{c_{earth}}{2\pi} + \frac{1}{2\pi} = \frac{d_{earth}}{2} + \frac{1}{2\pi}\)

So the radius of the ring equals the radius of the earth (3959 mi) plus 1/2 a yard divided by pi (5.73 in).

Using the metric system instead, the math would be the same, except that all dimensions would be in meters, so:

\(d_{earth} = 2 \times 6371\times 1000\)

If the circumference of the rope is increased by 1 m rather than 1 yd, the radius of the ring equals the radius of the earth (6371 km) plus 1/2 a meter divided by pi (15.9 cm).

 

To say it another way, I find it difficult to wrap my mind around this because the earth is so massive. How can adding a single yard of rope produce a ring with an area larger than 2 square miles?

Because the earth is so massive (or, in this case, so large in diameter).

Think about it another way. I have a rope loop that is in the shape of a rectangle that is 12,000 miles long on the long side and one inch long on the short side. So the total length of the rope is 24,000.0000316 miles and it encloses an area slightly larger than ten million square feet. I now add one inch of rope to each of the short sides making the total length of the rope 24,000.0000631 miles but now it encloses an additional ten million square feet. The reason is very easy to visualize -- the additional area that is enclosed is NOT due to the additional two inches, it's due to the original 24,000 miles of rope! All the two inches did was create a tiny increase in the separation of the two long sections of rope.

 

Because the earth is so massive (or, in this case, so large in diameter).

Think about it another way. I have a rope loop that is in the shape of a rectangle that is 12,000 miles long on the long side and one inch long on the short side. So the total length of the rope is 24,000.0000316 miles and it encloses an area slightly larger than ten million square feet. I now add one inch of rope to each of the short sides making the total length of the rope 24,000.0000631 miles but now it encloses an additional ten million square feet. The reason is very easy to visualize -- the additional area that is enclosed is NOT due to the additional two inches, it's due to the original 24,000 miles of rope! All the two inches did was create a tiny increase in the separation of the two long sections of rope.

@WBahn: Thanks, that was a helpful example.

I understand now how a tiny change in length at one place can cause a huge change somewhere else, but I still can't visualize how that would work with a loop of rope.

@paulktreg: Does my math make sense? I edited my previous post to include an explanation using the metric system.

 

Hi,

We can look at this in at least two different ways.

First, what we are after is the difference in diameters (or radii) D2-D1, where D1 is the diameter of the Earth and D2 is the diameter of the rope AFTER an increase of say 1 meter. With dimensions in meters, we get the result in meters:
d=D2-D1

Now since the diameter of the Earth is D1 and the circumference is C1=D1*pi, that means we can express D1 as:
D1=C1/pi

Since the diameter of the rope follows the same rule, we can state C2 as the circumference of the rope:
C2=D2*pi

or:
D2=C2/pi

We also know that C2 is 1 meter longer than C1, and working in meters we can express D2 as:
D2=(C1+1)/pi

So now finding the difference we have:
d=D2-D1=(C1+1)/pi-C1/pi

Factoring the right most side, we get:
1/pi

which is the increase in diameter, and since it is in meters this is about 0.318 meters or 318 millimeters. The radius is half that.


Another way to look at it is simply the rate at which the diameter (or radius) changes with a change in circumference.
Since we have any diameter D is equal to the circumference C divided by pi, we have:
D=C/pi

and taking the derivative of the right we get:
dD/dC=1/pi

so the slope of the change in diameter for a change in circumference is 1/pi. So this gives us equation;
dD=dC/pi

or in words, the change in diameter is equal to the change in circumference divided by pi. For a change equal to 1 meter we of course again get 1/pi, and again the change in radius is half of this.


This is an example that shows that intuition is not always reliable, especially when there is a concise formula available.

 

Hi,

We can look at this in at least two different ways.

First, what we are after is the difference in diameters (or radii) D2-D1, where D1 is the diameter of the Earth and D2 is the diameter of the rope AFTER an increase of say 1 meter. With dimensions in meters, we get the result in meters:
d=D2-D1

Now since the diameter of the Earth is D1 and the circumference is C1=D1*pi, that means we can express D1 as:
D1=C1/pi

Since the diameter of the rope follows the same rule, we can state C2 as the circumference of the rope:
C2=D2*pi

or:
D2=C2/pi

We also know that C2 is 1 meter longer than C1, and working in meters we can express D2 as:
D2=(C1+1)/pi

So now finding the difference we have:
d=D2-D1=(C1+1)/pi-C1/pi

Factoring the right most side, we get:
1/pi

which is the increase in diameter, and since it is in meters this is about 0.318 meters or 318 millimeters. The radius is half that.


Another way to look at it is simply the rate at which the diameter (or radius) changes with a change in circumference.
Since we have any diameter D is equal to the circumference C divided by pi, we have:
D=C/pi

and taking the derivative of the right we get:
dD/dC=1/pi

so the slope of the change in diameter for a change in circumference is 1/pi. So this gives us equation;
dD=dC/pi

or in words, the change in diameter is equal to the change in circumference divided by pi. For a change equal to 1 meter we of course again get 1/pi, and again the change in radius is half of this.


This is an example that shows that intuition is not always reliable, especially when there is a concise formula available.

A similar example is a railroad track 2 miles long. how much would the track length have to increase if the very center of the track was lifted by 1 millimeter?

@MrAl: Your explanation is probably good, but regrettably I can't understand it because I haven't taken calculus yet.

 

Hi,

That's ok, i think i quoted the wrong 'other' example anyway
That other example was an example of how numerical accuracy affects our calculations i think.

But the calculus part is correct, and the first part is just a factoring of the difference equation.

 

Hi,

That's ok, i think i quoted the wrong 'other' example anyway
That other example was an example of how numerical accuracy affects our calculations i think.

But the calculus part is correct, and the first part is just a factoring of the difference equation.

@MrAl: Sorry if I'm misunderstanding something, but it looks to me like the first part of your explanation in post #9 is basically repeating what I stated in post #6.

 

@MrAl: Sorry if I'm misunderstanding something, but it looks to me like the first part of your explanation in post #9 is basically repeating what I stated in post #6.

Hi,

Well it is almost the same, but my solution is pure algebra while yours is a mix of algebra and numerical calculations. I wanted to show the purely algebraic solution because that is a broader generalization of the solution. In my solution we arrive at a constant even though everything else is algebraic except for the increment, which i could have shown algebraically too but i didnt want to complicate it too much.
It is good that you noticed the similarities too though.

The calculus used isnt too hard to understand really. In that solution we only have to concentrate on the change that is occurring rather than any of the absolute quantities. This works for a lot of problems. And keep in mind that the 'change' in a variable is just like the 'increment', so the change in circumference is just 1 meter and that might be written as "dC", where the C stands for circumference and the lower case 'd' just stands for 'delta' or 'a small change' in the variable that follows it. The change in C causes a change in D (or R the radius) so the change in D we can write as "dD" which again just means 'a small change' in D. The small changes are related so we can find a formula like that. I am sure you will find this interesting if you look up more info on this on the web. I invite you to try. One of the general categories this would fall under is usually known as "Related Rates".

 

While this boils down to the same explanation, perhaps coming at it from the other direction might help.

If we have a circle of radius R, then we know that the circumference is C = 2piR

The question we want to explore can be asked two equivalent ways: How much does the radius increase for a given increase in circumference, or, how much does the circumference increase for a given increase in radius? Let's look at the latter and perhaps you will be able to visualize it better.

Let's call the original circumference C1:

C1 = 2piR

If we increase the radius to (R+ΔR), then the circumference becomes
C2 = 2pi(R+ΔR) = 2piR + 2piΔR = C1 + 2piΔR

The change in the circumference is

ΔC = (C2-C1) = 2piΔR

Another way of looking at it is to note that the circumference of a circle is roughly 6 times (i.e., 2pi) its radius. It doesn't matter how big or small the circle is, the circumference is roughly 6 times the radius. So if one circle has twice the circumference of another, then the radius of the first is twice the radius of the second. Similarly, if one circle has a circumference that is 1.00000001 times the circumference of the other, then the radius of the first is 1.00000001 times the radius of the second. Increasing the length of the rope is making a tiny change (one yard out of 24,000 miles) in the circumference of the circle and this is accompanied by the same fractional change (six inches out of 4,000 miles) in the radius.

 

While this boils down to the same explanation, perhaps coming at it from the other direction might help.

If we have a circle of radius R, then we know that the circumference is C = 2piR

The question we want to explore can be asked two equivalent ways: How much does the radius increase for a given increase in circumference, or, how much does the circumference increase for a given increase in radius? Let's look at the latter and perhaps you will be able to visualize it better.

Let's call the original circumference C1:

C1 = 2piR

If we increase the radius to (R+ΔR), then the circumference becomes
C2 = 2pi(R+ΔR) = 2piR + 2piΔR = C1 + 2piΔR

The change in the circumference is

ΔC = (C2-C1) = 2piΔR

Another way of looking at it is to note that the circumference of a circle is roughly 6 times (i.e., 2pi) its radius. It doesn't matter how big or small the circle is, the circumference is roughly 6 times the radius. So if one circle has twice the circumference of another, then the radius of the first is twice the radius of the second. Similarly, if one circle has a circumference that is 1.00000001 times the circumference of the other, then the radius of the first is 1.00000001 times the radius of the second. Increasing the length of the rope is making a tiny change (one yard out of 24,000 miles) in the circumference of the circle and this is accompanied by the same fractional change (six inches out of 4,000 miles) in the radius.

@WBahn: That makes sense, thank you! The explanation is so simple that it amazes me I didn't think of it before.

I found it interesting to note that if the circumference of a rope wrapped around any spherical object (not just the earth, but even a tiny marble) is increased by 1 yd (~1 m), the radius of the ring will always equal the radius of the object plus 5.73 in (~15.9 cm).

 

Think about how hard it is to tie a piece of string tightly around something like a tree. In fact try it, hint don't use a hitch. Then imagine scaling that up to the earth.

 

Think about how hard it is to tie a piece of string tightly around something like a tree. In fact try it, hint don't use a hitch. Then imagine scaling that up to the earth.

It's obviously just a theoretical problem, but I still find it interesting. I certainly don't expect to hear anytime soon of NASA having a satellite out in space tie a rope around the earth!

 

To summarize what WBahn said, change in radius varies directly with change in circumference:

\(\frac{c_2}{c_1} = \frac{r_2}{r_1}\)

So it follows that:

\(r = \frac{c}{2\pi}\\
\frac{c + 1}{c} = \frac{r + \frac{1}{2\pi}}{r}\)

And this is not only true for the earth, but also for any spherical object!

So now I understand why the radius increases by 5.73 in. regardless of whether the object the rope is wrapped around is the size of the earth or the size of a marble. In both cases, when the circumference is increased by a factor of x, the radius is as well. The only difference is that if the object is the earth, the percentage of increase is much, much less than it is for a marble.

 

I just came across an interesting resource that discusses this puzzle in depth: http://mathforum.org/mathimages/index.php/Rope_around_the_Earth

 

Yesterday I recalled an interesting brainteaser that I saw in a book a few years ago. It went something like this: Suppose a rope was tightly wrapped around the equator (picture A).
View attachment 80984
Then suppose that the length of the rope was increased by only a single yard. A mouse would be able to fit between the rope and the earth not just at one point (picture B), but at any given point around the earth (picture C). By working out the math, I could prove that the radius of the loop of rope increases by 5.73 (18/pi) inches. Doesn't this seem contrary to what one would expect?

Yes, I agree with you that 5 inches is a much bigger number than I would have estimated. Since I would have guessed that the 3 additional feet would have made little if any measure able difference. And, by the in-depth calculations and explanations, I think most other people are a bit startled as well. Nice and interesting post.