Maths help two :)

[Orionaut]

The other thread just reminded me of a question I saw many years ago (In New Scientist IIRC)

Consider the situation...

Imagine a string wrapped all the way around the Earths equator.

Now imagine a second string raised to a point 1M above the earths equator.

Q: How much longer is the second string than the first.

I am not searching for the answer here, simply the method.

What was your solution, and how did you arrive at it?

 

[Orionaut]

(2*pie*[r+1]) - (2*pie*r)

Where r is the average radius of the earth in metres at the equator.

But if you want to lay the string flat against every nook and cranny of the earth to get a true measurement, good luck

Correct, but too complicated.

There is a simpler way of looking at it

Thats what I am looking for, not the solution but rather the logic.

 

[Orionaut]

If you want to demonstrate working from first principle maybe but once you know that all the original figures drop out, it means you can happily calculate the change in one of those numbers simply by knowing the change in the other.

All the information required is in the question.

 

[Orionaut]

I was trying to post in a way that didn't just kill the thread half a dozen posts in

True

That was my logic when I first saw it. I haven't been told what the earths radius is, therefore it doesn't matter

Once you accept that, the answer is clear.

It is a bit like the "Monty hall" problem.

People feel the need to come up with all sorts of complex arguments to explain how this, apparently, counter intuitive situation can arise.

In fact it isn't complex at all.

The solution is that it has little to do with probability and everything to do with the fact that the game show host knows which doors have the car and which have the Goat!