Wired Equator

[Guest]

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.

[Guest]

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.

[Guest]

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!

[Guest]

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.

[Guest]

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.

[Guest]

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.

10=2*3.1416*X

X=10/6.2832

X=1.6

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.

[Guest]

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?

[Guest]

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...

[rookie1ja]

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

[Guest]

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

[Guest]

Hi,

Here is how I worked.

40000km=40000000m

2*pi*r1=40000000

r1=40000000/6.28

r1=6369426.75

NOW,

2*p*r2=40000010

r2=40000010/6.28

r2=6369428.34

r2-r1=1.59=1.6 approx

(Thought I would give steps mathematically in detail!!!)

[Guest]

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!

BoilingOil

[Guest]

or you could just swim under it...lol

[Guest]

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)

[Guest]

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.

[Guest]

To add to schmiggen's viable alternatives, without doing math, why not simply dig a hole where said string along the great circle matches said perfectly spherical planet and crawl under?

Nice math for the win though.

[Guest]

Easily.

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

[Guest]

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.

[Guest]

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.

[Guest]

The answer trail to this reads like the Monty Python skit where they try to determine whether or not a woman is a witch!

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.

[rookie1ja]

to recary:

word it

[Guest]

I enjoyed this problem... it seems to defy logic. I did the math again for myself, but it is still hard to believe adding only 10 meters to the wire would evenly lift it more than five feet from the surface of the "earth" sphere. Thanks.

[Guest]

Quick thought - 10m extra will form an "opening" 3.333r x 3.333r x 3.333r if you use it only one place on the equator, assuming the equator is were it is meant to gbe applied.

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

[Guest]

Depends. THe Wire at the time of measurement might be smaller than its size when it expands from the suns heat. Otherwise, no none of them could because it would be 10/40000 meters wide.

[Guest]

wow. you guys are thinking way too much on this one. no math is necessary. no geometry. look at it this way...

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.