As you may recall my mentioning on occasion, I'm a member of the "How Things Work" yahoo newsgroup. This group involves an on-going discussion that bounces from topic to topic with the agility of a mountain goat. Recently, someone on the group mentioned that something had reminded them of an "old chestnut" problem as follows:

"I have a mighty ball of string, 40 million metres of it, that reaches right round the world.

Now, by a Magick I shall not disclose here, I lift the string a metre off the ground (and sea), all the way round.

How much extra string will I need to close the gap?"

Actually, let's take a step back, because the "40 million meters" mentioned above is simply a close approximation to the actual circumference of the Earth. In order to make sure that we're all starting from a level playing field, let's say our piece of string is wrapped around the equator and that that the equatorial diameter of the Earth is 12,760 km (the polar diameter is 40 km less at 12,720 km).

Let's also assume that the surface of the Earth is perfectly smooth (no mountains or anything). So, if we were to magically lift our piece of string 1 meter off the ground, how much extra string would we need to "fill the gap"? Email me your thoughts and I'll post a compendium of answers in a few days time.

Questions? Comments? Feel free to email me – Clive "Max" Maxfield – at max@techbites.com). And, of course, if you haven't already done so, don't forget to Sign Up for our weekly Programmable Logic DesignLine Newsletter.

It would be interresting how this equation looks, which gives the result how far we have to move to north or south...
A real topic for my math teacher, who teased me with such questions back in school some decades ago.

If we assume a perfect circle of string then we can calculate the Diameter D
C=Pi x D
D=C/Pi = 40x10^6/Pi
When you lift the string by 1 Meter, the diameter increases by 2 Meters.
Then all you need to do is calculate the new circumference C and take the difference between old and new.
NewC=(C/Pi + 2) x Pi
To get the difference:
Difference = NewC ? C
= [(C/Pi + 2) x Pi ] ? C.
= C + 2xPi ? C = 2xPi
The answer I get is 2xPi
Rick.

If string is lifted 1m, effective diameter increases by 2m. let d = 2meters. If D is original diameter then difference in circumference is
C = (Pi*(D+d) - Pi*D) = Pi*d = 2*Pi meters = 6.283 meters

When I first saw this, I was surprised that the answer is independent of the starting circumference. It's interesting that the answer is surprising, since I'm not surprised that circumference is proportional to radius. Maybe that means that the distributive property of multiplication is nonintuitive. I doubt the explanation is that simple, since another form of the question doesn't seem to be a riddle at all: A company pays each of its employees $6.28 per hour, for a total of $40,000 per hour. How much more would the company need to spend per hour to hire one more employee?
C = 2*pi*R
R' = R+1
C' = 2*pi*(R+1) = 2*pi*R + 2*pi = C + 2*pi
Paul

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