- 17 Aug '05 17:22 / 1 editNot sure if this was posted already, but here goes nothin':

Imagine, if you would, a belt wrapped around the Earth along the equator. For the purposes of this puzzle, we'll consider the Earth to be a perfect sphere with a radius of 6371 km. Now:

(1) What is the length of the belt?

(2) If you added 1 km to the length of the belt, and pulled on the belt so that it was taught around the Earth again, how high off the ground would the belt be at the point of pulling?

(EDIT: Changed 1 metre to 1 km in part 2) - 17 Aug '05 17:30

1) pi*r sqrd = circumference so it would be : 400,452,567.938 kms*Originally posted by PBE6***Not sure if this was posted already, but here goes nothin':**

Imagine, if you would, a belt wrapped around the Earth along the equator. For the purposes of this puzzle, we'll consider the Earth to be a perfect sphere with a radius of 6371 km. Now:

(1) What is the length of the belt?

(2) If you added 1 metre to the length of the belt, and pulled on th ...[text shortened]... aught around the Earth again, how high off the ground would the belt be at the point of pulling?

that seems horribly wrong but its what the calculator spewed out. - 17 Aug '05 17:34

Yep, that's too big. It looks like you used the formula for the area of a circle with radius 6371 instead of the circumference.*Originally posted by Freddie2004***1) pi*r sqrd = circumference so it would be : 400,452,567.938 kms**

that seems horribly wrong but its what the calculator spewed out. - 17 Aug '05 17:46

wouldn't b) not be simply 1000m/(2pi)=159m?*Originally posted by Coconut***40031.140 = new belt length = L**

2(pi)r = L

L/(2(pi)) = r = 6371.000035014117055376417673853

r - earth's r = .000035014117055376417673853

so .000035 km off the ground. or .035 meters. or 3.5 centimeters.

I'm using 3.14159 for pi. - 17 Aug '05 17:53

That would be the height off the ground of a belt that was hovering equidistant over the Earth. I was thinking more like hanging the belt on a hook, and letting the Earth hang in the belt like a hammock (or like an ice cream cone).*Originally posted by Mephisto2***wouldn't b) not be simply 1000m/(2pi)=159m?**

On its side, it would look kinda like this: <O

- 17 Aug '05 17:53

im not sure if i got it right. is the belt still in shape of a circle then, that has the same height above earth everywhere? if yes id go like this:*Originally posted by PBE6***(2) If you added 1 km to the length of the belt, and pulled on the belt so that it was taught around the Earth again, how high off the ground would the belt be at the point of pulling?**

(EDIT: Changed 1 metre to 1 km in part 2)

2*pi*r = 40030.174 km + 1 km (result taken from part 1)

r = 6371.159 km

so its 159 m above earth. ummmm thats quite alot, it must be wrong, though i keep getting the same result.

you can get the result also without knowing the original r of earth:

1km/(2*ip) = 0.159 km

but im still not sure if the question was meant like that, because you were talking of a point of pulling. i guess my english is just too poor to understand it - 17 Aug '05 18:07 / 1 edit

oops.*Originally posted by PBE6***That would be the height off the ground of a belt that was hovering equidistant over the Earth. I was thinking more like hanging the belt on a hook, and letting the Earth hang in the belt like a hammock (or like an ice cream cone).**

On its side, it would look kinda like this: <O

wonder why my first calculation was so far off for the equidistance. - 17 Aug '05 18:13

No, it's not in the shape of a circle anymore. When you pull on the slack part, it makes a sharp point.*Originally posted by crazyblue***im not sure if i got it right. is the belt still in shape of a circle then, that has the same height above earth everywhere? if yes id go like this:**

2*pi*r = 40030.174 km + 1 km (result taken from part 1)

r = 6371.159 km

so its 159 m above earth. ummmm thats quite alot, it must be wrong, though i keep getting the same result.

you can get the re ...[text shortened]... you were talking of a point of pulling. i guess my english is just too poor to understand it - 18 Aug '05 16:19 / 1 edit

Well, if the belt was hovering equidistant above the equator the height off the ground is something like 0.159 km. When you pull the belt up so it touches the other side of the Earth, that doubles the height right off the bat. Then there's all that extra slack from the 40,000+ km of belt running around the sides.*Originally posted by Coconut***how the frack does it get so high if you only add 1 to the whole length?**

Another way to think about it is that letting the belt hover equidistant above the Earth is like plotting the average over a domain, and pulling the belt tight is like pulling up the middle of the graph and pushing down the sides to maintain the same area.