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Wired Equator

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#31 Garrek99


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Posted 03 August 2007 - 01:27 AM

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
radius = 40,000.01km / (2*3.1415926535)
This gives a radius of 6366.1993154 km or 6366199.93 meters

So with 40,000 km of wire, the radius is 636619.77 meters
So with 40,000.01 km of wire, the radius is 636619.93 meters.
The difference is .16 meters, or .52 feet. So there would only be about 6 inches clearance to walk through.

To avoid retracing all your steps it should suffice to say that "delta" 10 in circumference will result in "delta" 1.6 in radius. You remember "delta", it means "change".
The equation is:
C=2*Pi*R therefore for every 10 units of C (circumference) you would need to feed the equation 1.6 units of R (radius) to equalize both sides.

Now, you gotta figure out where you made the ten-fold mistake. You got 0.16 in the place of 1.6.
Oh, I see where the mistake took place. It's not 636619.77 meters it is supposed to be 6366197.7 meters.
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#32 papamike74



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Posted 24 August 2007 - 06:54 AM

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.

While all of this is true and accurate, one has to wonder: where are the parents, to allow the children to walk around wires all over the globe?
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#33 bennowak



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Posted 06 September 2007 - 12:42 AM

Now that we have all had a math lesson, has it occured to ANYONE that it never states the wire has to be around the equator? Place that same ring anywhere near the pole and you ha all sorts of room...
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#34 rookie1ja


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Posted 06 September 2007 - 10:39 AM

Now that we have all had a math lesson, has it occured to ANYONE that it never states the wire has to be around the equator? Place that same ring anywhere near the pole and you ha all sorts of room...

well, title of the puzzle is as follows: Wired Equator
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Posted 09 September 2007 - 09:32 PM

i really liked this one. cool concept of how minor things theoreticall can make huge changes mathmatically in the long run.

ran some extra numbers:
c® = 2 * pi * r (circumference given radius)
r(d) = c(d) / 2 / pi (radius given circumference)
a® = (pi * r) ^ 2 (area given radius)

then with the 40,000km and 10m example:
r(40 000 010) - r(40 000 000) = 1.59 m
-- 1.6 meters extra all the way around

and the really cool one (to me)
a(r(40 000 010)) - a(r(40 000 000)) = 200 000 025 meters squared.

.... 200 MILLION square meters differnce in surface area of the 1.5 m high ring around a 40,000km sphere using a 40,000.01 m wire. TWENTY MILLION !!!!!!

oooo. what if it were concentric spheres ... volume difference would be:
v(r(40 000 010)) - v(r(40 000 000))
810569671781086.3 cubic meters ..... BOO YA
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#36 siria-leon



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Posted 24 September 2007 - 05:05 PM

Here is how I worked.
r2-r1=1.59=1.6 approx
(Thought I would give steps mathematically in detail!!!)
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#37 BoilingOil


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Posted 26 September 2007 - 01:18 AM

Wired Equator - Back to the Logic Puzzles
The circumference of the globe is approximately 40 000 km. If we made a circle of wire around the globe, that is only 10 metres longer than the circumference of earth, could a flea, a rabbit or even a man creep under it?

Wired Equator - solution
It is easy to subtract 2 equations (original perimeter = 2xPIxR, length of wire = 2xPIxR + 2xPIx(new R)) and find out that the result is 10m/(2xPI), which is about 1.6 m. So a smaller man can go under it and a bigger man ducks.

I’m confused. Are you saying that adding 10 Meters of wire to a perfectly measured 40,000 km of wire that fit snuggly around the planet would give you enough freed up space to elevate the wire 1.6 meters off the ground?? The way the question is worded is " If we made a circle of wire around the globe, that is only 10 meters longer than the circumference of earth" Does that mean you are then using 40,010 meters of wire??? I would think that 10 meters of wire compared to 40,000 would gain you such a microscopic amount it would be almost impossible to measure..

Sorry for the ignorance..

In that case, I'll confuse you even more!
The above is not entirely correct, but for a different reason: it assumes that the rope must be the same distance from the earth's surface all around the globe!
If on one end, we would let go off the rope, allowing it to fall to the ground, and on the other side we would pull the rope up as high as possible, an elephant could pass under it without hitting the rope! So there would not be a man tall enough to even need to bow!

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#38 hunter04535



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Posted 02 October 2007 - 02:39 AM

or you could just swim under it...lol
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#39 woodyallen



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Posted 04 October 2007 - 08:35 PM

okay - here's the calculation:

circumference = 2*pi*radius = 40,000 (divide by two below)
pi*radius = 20,000 (divide by pi below)
radius = 6366.19718 (original radius of earth tight wire)

new circumerence = 2*pi*radius = 40,000.01 (divide by two below)
pi*radius = 20,000.005 (divide by pi below)
radius = 6366.19877 (new radius of circle with +10 m circumference)

difference in radius = 6366.1987 - 6366.1971 = 0.0016 km

0.0016 km = 1.6 meters difference in radius (5 foot 3 inches)
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#40 commisioner98



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Posted 21 October 2007 - 11:52 PM

One could just as easily add no extra length to the wire, find a place where it crosses a shallow body of water and swim under it.
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