# That's a big belt

PBE6
Posers and Puzzles 17 Aug '05 17:22
1. PBE6
Bananarama
17 Aug '05 17:221 edit
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 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. 17 Aug '05 17:30
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?
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.
3. 17 Aug '05 17:341 edit
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.
pi*r^2 is area not circumference.

you want 2(pi)r = 40030.140 km

I'll look at part 2 in a second.
4. PBE6
Bananarama
17 Aug '05 17:34
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.
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.
5. 17 Aug '05 17:40
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.
6. 17 Aug '05 17:46
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.
wouldn't b) not be simply 1000m/(2pi)=159m?
7. PBE6
Bananarama
17 Aug '05 17:53
Originally posted by Mephisto2
wouldn't b) not be simply 1000m/(2pi)=159m?
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

8. 17 Aug '05 17:53
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)
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 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 ðŸ˜‰
9. 17 Aug '05 18:071 edit
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

oops.

wonder why my first calculation was so far off for the equidistance.
10. PBE6
Bananarama
17 Aug '05 18:13
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 ðŸ˜‰
No, it's not in the shape of a circle anymore. When you pull on the slack part, it makes a sharp point.
11. Bowmann
Non-Subscriber
17 Aug '05 21:22
Cor, this is an old one!
12. 18 Aug '05 08:10
Originally posted by PBE6
No, it's not in the shape of a circle anymore. When you pull on the slack part, it makes a sharp point.
forgot the sharp point vs cirlce part.
Could that be 12.17 km above the surface? Would be amazing.
13. PBE6
Bananarama
18 Aug '05 14:37
Originally posted by Mephisto2
forgot the sharp point vs cirlce part.
Could that be 12.17 km above the surface? Would be amazing.
I got 12.15 km using Excel, so 12.17 km sounds right to me! As Bowmann pointed out, "Cor, this is an old one!" but it's an interesting puzzle nonetheless.
14. 18 Aug '05 15:25
Originally posted by PBE6
I got 12.15 km using Excel, so 12.17 km sounds right to me! As Bowmann pointed out, "Cor, this is an old one!" but it's an interesting puzzle nonetheless.
how the frack does it get so high if you only add 1 to the whole length?
15. PBE6
Bananarama
18 Aug '05 16:191 edit
Originally posted by Coconut
how the frack does it get so high if you only add 1 to the whole length?
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.

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.