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Imagine a smooth metal ring around earth’s equator, and imagine a metal cable wrapped around the metal ring so that nothing can squeeze between cable and the ring. Assume both the ring and cable are perfect circles, and cannot shrink or stretch. The cable has 3 feet added to its circumference and is floats off the ring so that the slack is equally distributed around earth.
Poll: What is largest object that will fit between cable and ring?G) a grapefruit (~5 inches diameter) (11) 29% B) a virus (~100nm) (8) 21% H) a pumpkin (~1.5 feet diameter) (7) 18% A) an electron (Less than 1x10-13 cm) (3) 8% C) single skin cell (~20 microns) (3) 8% D) a grain of sand (~200 microns) (3) 8% E) a typical pearl (~9mm diameter) (3) 8% F) a grape (~1 inch diameter) (0) 0% 38 total votes Your vote: What is largest object that will fit between cable and ring? (Vote): A) an electron (Less than 1x10-13 cm)
(Vote): B) a virus (~100nm)
(Vote): C) single skin cell (~20 microns)
(Vote): D) a grain of sand (~200 microns)
(Vote): E) a typical pearl (~9mm diameter)
(Vote): F) a grape (~1 inch diameter)
(Vote): G) a grapefruit (~5 inches diameter)
(Vote): H) a pumpkin (~1.5 feet diameter)
try to guess it before looking at answer or doing any calculations
+ Show Spoiler +the earth has a diameter of 41,865,458 feet and the gap has a size of 5.73 inches, approximately a grapefruit. It will be the same for any spherical object. This really blew my mind when I first heard it but i saw the proof and it works PROOF+ Show Spoiler +
the difference between cable’s circumference and rings circumference is 3 feet, so
2*Pi*r2 = 2*Pi*r1 + 3
2*Pi*r2 - 2*Pi*r1 = 3
2*Pi*(r2-r1) = 3
We are trying to find r2-r1, so
r2-r1 = 3/2*pi
which equals 0.477 feet or approximately 5.7 inches
 
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the minute you said feet i gave up
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If you solve the equation, you obtain gap = 3 feet / 2 Pi = 5.73 inches.
Incredible indeed. I would have said it was much smaller but decided to do some maths first 
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On August 18 2010 12:43 exeexe wrote:
the minute you said feet i gave up
3 feet isnt that hard to picture, it's approximately 90 cm
the general idea is to think of adding 1 meter to the circumference and thinking how much bigger the cable gets
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I don't understand any of this...
The only thing that blew my mind is how confusing this whole problem is.
So, is this saying the Earth is so smooth around the equator line that the standard deviation from the mean is only 5,73 inches? o.O
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On August 18 2010 12:48 Leath wrote:
I don't understand any of this...
The only thing that blew my mind is how confusing this whole problem is.
So, is this saying the Earth is so smooth around the equator line that the standard deviation from the mean is only 5,73 inches? o.O
no, the earth isn't really the point, the only role the earth plays is to give u an idea of how large the circle is; and you are assuming the circles are perfect circles
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voted pumpkin just because no one else did :D
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well since most of tl is hopefully not as backwards as the US when it comes to measuring. here's the metric proof
+ Show Spoiler +assume circumference of earth is 40 07600m, the ring has no thickness, and the ring and metal band is perfectly 40 07600m.
C1 = 40 076000m
C2= 40 076000.90m
c1=2pir1
r1=6378293.499m
r2=6378293.643m
d=r2-r1
d=14.3cm
therefore a 14.3 cm object will fit between the band and the ring
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Would have been fine if you either used all metric system or no metric system.
I think I calculated there should be a .14 meter gap, and then I had no idea which answer would match a .14 meter gap.
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On August 18 2010 12:59 Malgrif wrote:well since most of tl is hopefully not as backwards as the US when it comes to measuring. here's the metric proof + Show Spoiler +assume circumference of earth is 40 07600m, the ring has no thickness, and the ring and metal band is perfectly 40 07600m.
C1 = 40 076000m
C2= 40 076000.90m
c1=2pir1
r1=6378293.499m
r2=6378293.643m
d=r2-r1
d=14.3cm
therefore a 14.3 cm object will fit between the band and the ring
Or you could you just convert the final result
Since the rest of the proof is just pure letter variables.
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You don't have a correct answer in those options. The correct answer is nothing, because it's stated in the premise of the problem.
Imagine a smooth metal ring around earth’s equator, and imagine a metal cable wrapped around the metal ring so that nothing can squeeze between cable and the ring.
Right there.
Man, I love these math problems :> Got any more?
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Hehe, that is one of the few things I remember from my first semester on our university. Guess that shows how hardworking student I am T_T ...
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On August 18 2010 13:13 Sadistx wrote:You don't have a correct answer in those options. The correct answer is nothing, because it's stated in the premise of the problem. Show nested quote +Imagine a smooth metal ring around earth’s equator, and imagine a metal cable wrapped around the metal ring so that nothing can squeeze between cable and the ring. Right there. Man, I love these math problems :> Got any more?
The cable has 3 feet added to its circumference and is floats off the ring so that the slack is equally distributed around earth.
??
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On August 18 2010 13:13 Sadistx wrote:You don't have a correct answer in those options. The correct answer is nothing, because it's stated in the premise of the problem. Show nested quote +Imagine a smooth metal ring around earth’s equator, and imagine a metal cable wrapped around the metal ring so that nothing can squeeze between cable and the ring. Right there. Man, I love these math problems :> Got any more?
lol
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The real question is, why on earth is a cable floating three feet off of the surface of the earth?
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NOOOOOOOOOOOOOO I HAZ PUMPKIN
![[image loading]](http://www.norcalblogs.com/bumpkins/pumpkin1.jpg)
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I'd say a virus, because while the change is tiny, it's not so insignificant that an electron could fit but not so significant that a pearl could.
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On August 18 2010 13:38 zer0das wrote:
The real question is, why on earth is a cable floating three feet off of the surface of the earth?
A better version of the riddle would be that the cable is supported in the air to it's maximum circumference, rather than just floating XD
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On August 18 2010 15:20 AlienAlias wrote:Show nested quote +On August 18 2010 13:38 zer0das wrote:
The real question is, why on earth is a cable floating three feet off of the surface of the earth? A better version of the riddle would be that the cable is supported in the air to it's maximum circumference, rather than just floating XD
It's not floating 3 feet off... I admit the "nothing" in the premise was confusing, but I found the problem intersting as the final results is a lot different that what I expected.
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On August 18 2010 15:26 endy wrote:Show nested quote +On August 18 2010 15:20 AlienAlias wrote:On August 18 2010 13:38 zer0das wrote:
The real question is, why on earth is a cable floating three feet off of the surface of the earth? A better version of the riddle would be that the cable is supported in the air to it's maximum circumference, rather than just floating XD It's not floating 3 feet off... I admit the "nothing" in the premise was confusing, but I found the problem intersting as the final results is a lot different that what I expected.
Yeah, it's definitely one of my favorite teasers. Realistically, you'd think that adding a little bit of length to a rope that goes around the world would do nothing, considering how massive the world is, but when you think about it analytically the circumference and radius relate directly at a certain ratio (c = (pi)2r) so it doesn't matter what the previous radius was.
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EPT
