Wired Equator

[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?

the way it is worded it sounds to me like the extra 10m of wire wouldnt make any difference.. it would just wrap around the begining of the wire.. i that makes sence

but i understand it now... but the wording isnt great

[Guest]

I have nothing to contribute that hasn't already been said...

...but I am amused at the sight of how many people confused the measurements, even after a full year has passed from the original posting, making the reference to the 40,000 KILOMETERS as 40,000 METERS...

>>> Many of you might agree that THAT had to be the real Brain Teaser here!!!! <<<

...

if a wire with a length of 40010 meters is stretched out in an equatorial line around the sphere...

...

Sorry for picking on you specifically Ciahra... the number you wanted to quote was intended to be 40,000,010 meters.

(I know... it does not change anything to the comments made...)

[Guest]

Unbelievable as it may seem, but it is so. Radius of circle made of 40,000 km and 10 meters long wire is just about 1.6 meters longer than the radius of circle made of 40,000 km long wire.

But feel free to prove the opposite.

Who said anything about radius? If you have 40,000 km of wire and add 10 meters of wire to it. You now have a wire that is 10 meters longer than it was. If you secure the 40,000km wire to the earth at two points, lets say one meter apart. cut the 40,000 km wire. splice in the 10 meter section of wire at this point. Then anything that would fit through a 10 meter loop of wire would fit under the wire.

[Guest]

This one is easy!

If C/D=pi then C/pi=D

So 40,000/3.141=12734.797

And 40,010/3.141=12734.981

So the diametre of earth is 12734.797km and diametre of rope around the earth is 12734.981km so the difference is ap. 200m. That is on both sides: (O) so half is 100m (O above the ground approx.

So a man could easy walk under, am I right?

[Guest]

Though the person with the full calculation had one of the most intelligent answers posted, the formula is wrong.

The formual is C=2πr, therefore r=C/2π, not as they had it r=(Cπ)/2

Therefore the correct answer is 1.6m (rounded) as the solution stated if the Earth is assumed to be a perfect sphere. And to explain a little more, this equation finds the radius of the Earth with the original wire length of 40,000,000m where r turns out to be 6,363,650.83m. If 10m is then added to this very long wire, the new circumference will be 40,000,010m and r will be 6,363,652.42m. Therefore, the distance between the new radius and the old radius will give you approximately 1.6m. And all those of us who are under 1.6m tall can go under the new wire

I'd like to make another comment about the measures - you can't add 10m to 40000km and get 40010km. FYI 1km=1000m. So many people got it wrong...

[Guest]

OK, if it looks like a math problem and smells like a math problem then it must be a math problem, right?

Well, what if it isn't a math problem after all, but rather a word problem.

When I think of a "globe" I don't think of the actual physical Earth, but rather a model of the Earth. Certainly (though I haven't checked lately) the circumference of a "globe" may very well represent 40,000 km at the equator. The question then asks to take a piece of wire and wrap it around the "globe". However, the length of the wire is now to be 10 meters longer then the circumference of the EARTH. Our brains assume that "globe" and Earth are one in the same. But, the question does reference two distinctly different entities. One being a "globe" and the other being the "Earth". So if the run-of-the-mill globe is somewhere around 18 inches in diameter and you take a wire that is the actual circumference of the Earth plus 10 meters in length and wrap it around the "globe", you could fit the entire Earth under it.

Seemingly this is not the intent of the question, but then again, why couldn't it be? Could the use of the words "globe" and "Earth" have been deliberately chosen to misdirect the reader?

[Guest]

Isn't the circumference determined at sea-level? Has anyone considered that a wire even a kilometer longer than the circumference of the earth WOULD NOT make it around the equator due to the altitude of the mountains in both Africa and South America?

[Guest]

Interestingly enough, it doesn't matter what the original diameter of the object is. If you add 10 meteres to the original circumference x, the change in radius will be 1.6 meters.

Here's the algebra: Solving for change in radius (R2-R1)

Knowns:

x1 = 2*Pi*R1

x2 = 2*Pi*R2

x1 = x2 - 10

Solution:

R1 = x1 / (2*Pi)

R2 = x2 / (2*Pi)

so by substitution, R2 - R1 = x2 / (2*pi) - x1 / (2*pi)

-or-

R2 - R1 = (x2 - x1) / (2*Pi)

then

R2 - R1 = [x2 - (x2 - 10)] / (2*Pi)

The x2's cancel out. The result, regardless of x2 or x1 is simply:

R2 - R1 = 10 / (2*pi) or 1.6

Crazy to think that whether the object is as small as marble, or as big as the galaxy if you add 10 to the circumference measuring tape, you can pass a small child underneath anywhere.

Edited April 17, 2008 by Kenitzka

[Guest]

yes see if you can find a piece of wire that just so happens to be that long, then honey feel free to try, but I'm not going to answer the question because of the wording but if I had to I would probably say no. And if you ever do decide to try that give me a call because that would be something that I would most definitely love to watch.

[Guest]

it is a very easy one

r= C / (2pi)

so D= 2r

so the D1 of the 40,000 km is:

r= 40000 / (2*3.141592)

r= 40000 / 6.283184

r= 6366.19904812

D1= 12,732.3980962 km

and the D2 of 40,000.01 is

r= 40000.01 * (2*3.141592)

r= 40000.01 * 6.283184

r= 6366.20063965

D2= 12,732.4012793 km

Susbtract

D2-D1

12,732.4012793 -12,732.3980962 = 0.0031831 km = 3.183 meters

But beacuse the it is 3.183 m on both sides of the eart it will leave 1.59m per side

[Guest]

Wouldn't a flea jump over, rather than creep under the wire?

[Guest]

That makes no sense. Everyone is making it harder than it really is. Imagine the wire laying on the floor right in front of you running perfectly across the ground. Now pick a spot on that wire and cut it. Add 10M of more wire. Problem solved, now you can easily walk under it.

[Guest]

The 10 meters isn't spread equally around the globe necessarily. If it were then the distance between the globe and the wire would be very small indeed. I'm sure one of these super brains could easily figure the distance if that were a condition.

[Guest]

Well, the way I make it is

C= 40000000M therefore (Cx3.142857143/2=r) the Radius is 62857142.86 M

C= 40000000 + 10 is C = 40000010M therefore the radius is 62857158.57 M

62857142.86 - 62857158.57 is 15.71428571M.

Your answer of 1.6 M would be correct if the wire were only 1 M longer.

Or have I missed it completely?

the formula u used to calculating radius was r=C*(pi)/2

check the maths book dude its C/2*(pi)

[Guest]

Wouldn't a flea jump over, rather than creep under the wire?

lol that's what my cousin said

[Guest]

I have seen a very similar question to this....I really like this and found it very interesting.

This was the question I saw.

Lord Fimbleton lost 6 meters of his cricket fence, how much shorter would you expect the boundary, which was 50m away from the centre?

Only 1 meter shorter. So the radius is now 49 meters

Now the Gods also hold a cricket match, a very vast cricket ground, billions of miles across, but they have also lost 6 meters of fencing, how much shorter would their ground be?

1 meter. So the radius will be 1 meter shorter.

You would think that, for the second one, it would be less, but it isn't

This is why:

We know that the perimeter for a circle is P= 2pi*r

Lets do this simultaneously

Old P=2pi*Old r 1

New P=2pi*New r 2

2-1

New P-Old P=2pi*(New r-Old r)

6=2pi*(New r-Old r)

6/(2pi) = New r-Old r

which is approx. 1 meter. So the numbers don't matter. Doesn't matter how big or small it was to begin with. Don't let that deceive you.

I recommend to anyone who enjoys these, or similar, puzzles to buy and read 'Why do buses come in threes?' by Rob Eastway and Jeremy Wyndham. And their other books too. They are very good, interesting and enjoyable to read.

[Guest]

Interestingly enough, it doesn't matter what the original diameter of the object is. If you add 10 meteres to the original circumference x, the change in radius will be 1.6 meters.

Here's the algebra: Solving for change in radius (R2-R1)

Knowns:

x1 = 2*Pi*R1

x2 = 2*Pi*R2

x1 = x2 - 10

Solution:

R1 = x1 / (2*Pi)

R2 = x2 / (2*Pi)

so by substitution, R2 - R1 = x2 / (2*pi) - x1 / (2*pi)

-or-

R2 - R1 = (x2 - x1) / (2*Pi)

then

R2 - R1 = [x2 - (x2 - 10)] / (2*Pi)

The x2's cancel out. The result, regardless of x2 or x1 is simply:

R2 - R1 = 10 / (2*pi) or 1.6

Crazy to think that whether the object is as small as marble, or as big as the galaxy if you add 10 to the circumference measuring tape, you can pass a small child underneath anywhere.

I know. It's quite mind blowing to think that. I posted a similar thing (I didn't look through all the posts). I saw this in a book (Why do buses come in threes) and it just goes to show just how wrong gut feelings can be, and how logic and maths can show it.

[Guest]

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

LOL! But yep...exactly. I also stated (and someone else did too) a similar illusion. Where you can remove or add the same amount to both a small and large object and still have the same difference.

e.g. a tennis ball(A) and tennis ball(B) being 7 cm bigger in circumference and a planet(A) and planet(B) being 7 cm bigger in circumference. They will both have the same difference in radius, but you wouldn't expect that as you would expect the planet to have a smaller change, because it's a bigger object.

What people need to remember is that the radius will increase/decrease by the same amount, regardless of the size of the object, BUT the change will be less significant in the larger object, as it was already huge.