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Wednesday, August 4, 2010

Circumference Classic

I've adapted this classic anti-intuitive puzzle from one of Clifford Pickover's renditions of it:

Imagine a rope tightly encircling the 'equator' of a basketball. How much longer would you have to make the rope for it to now be (when stretched out as a circle) one foot from the surface at all points?

Now, imagine a rope similarly around the equator of an Earth-sized sphere (around 25,000 miles long). How much longer would you now have to make that rope for it also to be one foot off the ground all the way around the equator?

The AMAZING answer is 2π (or approximately 6.28) additional feet for BOTH the basketball and the Earth! If r is the radius of the Earth (in ft.), and 1 + r is therefore the radius in feet of the enlarged circle, we need only compare the rope circumference (length) before (2π r) and after [2π (1 + r)] (and ditto for the basketball!)

1 comment:

Alexander Bogomolny said...

A curious thing is that the area of the ring between the two (rope) circles only depends on a single parameter which is not the radius of either of the circles:

http://www.cut-the-knot.org/Curriculum/Geometry/PythFromRing.shtml