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# Wired Equator

### #41

Posted 04 December 2007 - 12:03 PM

Nice math for the win though.

### #42

Posted 08 January 2008 - 06:34 AM

**It never said that it was wrapped around the world, just that it was 10 meters longer than the circumference.**

### #43

Posted 11 January 2008 - 02:10 AM

Easily.

It never said that it was wrapped around the world, just that it was 10 meters longer than the circumference.

Maybe I'm a bit off on my thinking here, but it seems that would be plenty of room for even a small car to pass through. Wrap the wire around the earth, place two stakes in the ground a meter or so away from each other and pin the wire to ground at those two points, with the extra 10 M of slack between the two stakes. You now have a 1 meter gap in which the wire can be raised (use tent poles or hook to a tree branch to hold in place) atleast 4 meters off the ground (4 meters up, 1 meter over, and another 4 meters back down to the other stake, with an over an extra meter to spare). This would leave plenty of room to move under the wire. The question never says the wire is distributed evenly across the surface of the earth.

The other option would be to dig a hold underneath it, cross under the wire where it runs over a canyon, or possibly swim under it when it crosses the ocean or a lake.

### #44

Posted 17 February 2008 - 06:47 AM

Earth is not a smooth surface by any standards.

But still, those feel like cheating answers somehow.

Actually, it is. If you were to view the Earth on the same scale as, say, a billiard (pool) ball, its actually smoother than the ball!

Everything depends on your frame of reference.

### #45

Posted 17 February 2008 - 03:27 PM

This question is ambiguous from the start -"world is

*approximately*40000km?" This leaves room for the flea immediately.

But lets for a minute assume that we're talking about a perfect sphere, which some people have taken for granted and started in on the geometry principles....

Unfortunately given the wording of the question, the geometric principle only is worth using if you're assuming the 10m (about 33 feet) of extra cable is distributed evenly to make another ever so slightly larger perfect circle around the 40000km circle.

However, it doesn't say in the question that this must be so, so I'm assuming you can use all 10m (about 33 feet of slack) the slack to produce a deformity at one point in the sphere large enough to fit all thre contestents through. At the theoretical limit you could make an area 5m (15feet or so) high (you need the other 5m to complete the other half of the arc deformity) that has almost no width. At this point we should consider the width of the person...see why this question is dumb? Some simple integral calculus could provide you with the area beneath the arc deformity which would give you a better idea of the continuous space underneath through which the contestants would have to fit...but again to perform the integral we'd need the question to have provided a function to describe the arc.

Now, some practical implications of this might involve the tinsel strength of the wire, its pliability in forming the small arc deformity. The mass of the wire and strength required to cause the deformity, etc...but all this could fall into the "approximately" area that makes this question inherently confusing.

I think the question should be worded more clearly.

### #46

Posted 17 February 2008 - 04:37 PM

word it

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### #47

Posted 21 February 2008 - 10:32 PM

### #48

Posted 21 February 2008 - 10:48 PM

Sorry if this is a duplicate -too many answers to go through!

### #49

Posted 26 February 2008 - 01:27 AM

### #50

Posted 28 February 2008 - 07:17 AM

a sphere, perfect in nature, made of metal so as not to be cut, precisely 40k meters in circumference and with standard gravity (for you geeks out there...)

if a wire with a length of 40010 meters is stretched out in an equatorial line around the sphere, when the sphere is enveloped (along that line) there will be 10 meters in abundance. those 10 meters can be manipulated in such a way that many large things could pass under the wire, as the space created could, for example, be 4 meters high and 2 meters wide. now you can do the math.

using equations that involve circumference, diameter, and radius assume the wire would be equidistant from the sphere at all points around that sphere. such conditions were not given in the puzzle, so should not be inferred.

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