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  2. Now, tie a thread around earth's equator...

Now, tie a thread around earth's equator...

By thescian on July 4, 2007.

Imagine this: Tie a thread around the earth's equator. Then, say, you let the thread out by 2*Pi metres (Pi=Ï, the mathematical constant) to make a thread circle that stands slightly above the earth's equator. Now, how far above the surface of the equator will the thread stand? In other words, what's the difference between new radius of the thread circle and the original radius (original radius being the earth's equatorial radius)?

This question was posed by Ramanujan, the mathematical genius from India, to a young student he was tutoring in Madras. The answer is so counter-intutive that I jumped off my chair. Give it a go (simple math really).

Now, for those who already know this, can you explain to me why the result is so counter-intuitive? I just can't wrap my head around this although the math seems quite straightforward.

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So it hovers 1m above the surface ?
And it does whatever the radius of the ball is, start with a 1m ball and you get a 2m radius... If I'm not mistaken ?

I'd guess the confusion comes from the fact that seen from a human point of view, 1m is big, whereas compared to the initial radius in meters it is a tiny extra amount.

Weird, isn't it.
l=2*pi*r so
dl = 2*pi*dr

It seems that we intuitively expect
the height change dr to be proportional to the relative change in the length dl/l, but it ain't.

Perhaps it helps to think of the earth as being flat - then the length of the rope doesn't need to change as you move it up and down.

Our intuition tells us the height change should be proportional to the length change. But what is happening is the RADIUS change is proportional to the length (circumference) change.

The radius has changed by roughly 3 feet, added to the 4000 miles, while the circumference has changed by 18 feet, added to the 24,000 miles.

I have to admit that I don't see why this is surprising.... It sounds right to me. Change the circumference by 2pi meters, you change the radiums by 1 meter-- it's a circle, so that's what you'd expect.

I wouldn't expect dr to be proportional to dl/l... I'd expect dr/r to be proportional to dl/l.

Perhaps I've just been mucking about with circles too long. I suppose it is comforting that humans can train their intuition.

-Rob

Maybe it's counter-intuitive because the result would be very different in reality? The earth is not a perfect smooth sphere.

Sophie, yes, it hovers one meter above the surface.

It's the difference in the proportions that skewed my perception. I implicitly expected a great length to thread to be let out for it to stand a meter above the surface of a huge planet. 2 pi is so small...

Rob, you'll find a challenge for intuition in the english weather. :-)

Selva: maybe think of it like the disproportional speed that the cutting part of a pair of scissors moces as compared to the motion of the blades of the scissors. If you tried to imagine a pair of scissors as large as the radius of the earth, and you standing in the gap between the blades, the pro-rated tiny fraction of a human-scale added circumference would make the scissors move from already near-parallel to slightly less parallel, but the even this tiny change in scissor angle would move the scissor intersection significantly.

Following up on csrster's thoughts:

I find it most simple to take it a step further to think of the equation as

dr/dc = 1/(2*Pi)
or
change of radii per unit change of circumference = 1/(2*Pi)

So one unit change of the radius will increase the circumference by 1/(2*Pi). So naturally you can tell that an increase of the circumference by 2*Pi will result in radius change of 1 unit.

I get the same thing. Initially, the length of the thread is 2pi, then it becomes 2pi + 1. The distance from the center of the Earth, given the thread's length is equal to L/2pi. Since L = 2(pi)r + 2pi, we get the distance from the center to be r + 1. Thus the thread is one meter above the ground.