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# Wired Equator

67 replies to this topic

### #21 Veracity

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Posted 30 June 2007 - 12:21 PM

The illusion comes from the notion that 10 meters is infinitesimally small compared to 40,000 km (so the change should be really really small). Yes, so is 1.6 meters compared to 40,000/(2*PI) km

VERY VERY Well Said! That's what I said minus all the educated words..lol
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### #22 mannu

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Posted 06 July 2007 - 06:16 AM

I think there is another way to interpret the question. Its never stated that the extra 10 meters should be evenly distributed along the circumference. So 40000 Km of wire can be used to encircle the earth and u r left with 10 meters. That wd mean a height of about 5 meters (10/2). So i guess a man can surely pass thru it...
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### #23 stephcorbin

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Posted 08 July 2007 - 04:36 AM

I do agree with Mannu. 40 000km requires a wire of at least 40 000km. You hence cover the original circumference. Now at a very location you add 10 meters of wire: simply putting it on the ground, the wires will overlap.

If you want to maximize the eventual surface available, you take from the original wire 5 meters on the left and 5 on the right to have them stick perpendicular to the ground and use your 10 meters to complete the new figure.

At the end, you have approximately 5 x 10 meters of surface available. Well enough for whatever you may think. Indeed it's not precisely 50 sq/m; the earth being a rather spheric, it's a little less.

Distributing the additional 10 meters over the entire circumference would leave a gap of 1.6m. But I'm still wondering how this new wire will stand up by itself! Thus my more pragmatic interpretation of the problem...

This was my two cents.
Stephane
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Posted 09 July 2007 - 10:25 PM

No it does not mean having a wire of 40,010 meters but 40,000.01 km. But still, the 0.01km change leaves room for children to walk around.

Ahh... That was sneaky -- I missed the conversion. I didn't read that one was Km and the other was M.
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### #25 iprogroup

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Posted 10 July 2007 - 08:17 AM

Check the value of PI you have used. The correct value of PI is 3.1415926536. Use this and you will get the correct answer.
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### #26 iprogroup

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Posted 10 July 2007 - 08:20 AM

Check the value of PI you have used. The correct value of PI is 3.1415926536. Use this and you will get the correct answer.
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### #27 kabooms1fire

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Posted 17 July 2007 - 12:52 AM

actually the point is moot if you had a wire that snugly fit around the world assuming that there are no hills vallys or bodies of water (an imposibility) and you added 10M you could just lift the slack in a single place to form a triangel of almost 7m high. this is why i dont like mixing riddles with math or science. all the variables.
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### #28 han42931

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Posted 26 July 2007 - 03:31 PM

well if we think about a wire going around the circumference of a theoretically spherical planet, then we think about cuting the wire at any given spot and adding ten meters of wire to that spot, you could reasonably assume that the wire would go up 5 meters and down 5 meters at that specific spot... so you could create an opening of any shape that had a ten meter perimiter and have even a very large man walk through it!, you could make a 2.5meter square opening if you were in a ideal world, and have wire on all sides of the opening!
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### #29 Garrek99

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Posted 27 July 2007 - 01:13 AM

well if we think about a wire going around the circumference of a theoretically spherical planet, then we think about cuting the wire at any given spot and adding ten meters of wire to that spot, you could reasonably assume that the wire would go up 5 meters and down 5 meters at that specific spot... so you could create an opening of any shape that had a ten meter perimiter and have even a very large man walk through it!, you could make a 2.5meter square opening if you were in a ideal world, and have wire on all sides of the opening!

Guys, you gotta read the entire discussion before you post.
At least 2 people before you suggested the same thing. I think one was a rectangle and the other a triangle.
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### #30 evan123456

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Posted 02 August 2007 - 07:18 PM

I think we can assume we are dealing with a perfect sphere. The object of the question is to see if you add 10 meters to the wire, how much space will be under the larger circle. So if you have a wire tracing the circumference of the sphere, 40,000 kilometers long, it is forming a circle. The radius of a circle is r = circumference / (2*pi)

r = 40,000km / (2*3.1415926535)
This gives the circle a radius of 6366.1977239 km or 636619.77 meters

If you add 10 meters to the wire and wrap that around the same perfect sphere