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ThudnBlunder
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 Rope Around the Earth   « on: Apr 9th, 2003, 5:39pm » Quote Modify

Assume the Earth is a perfect sphere of radius r and suppose a rope of zero elasticity is tied tightly around it.
One metre is now added to the rope's length. If the rope is now pulled at one point as high as possible above the Earth's surface, what height will be reached?

 « Last Edit: Apr 10th, 2003, 9:59am by ThudnBlunder » IP Logged

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Icarus
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 Re: Rope Around the Earth   « Reply #1 on: Apr 9th, 2003, 8:12pm » Quote Modify

For a 40,000 Km circumference of the Earth, I get ~193 meters (first-order approximation)
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aero_guy
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 Re: Rope Around the Earth   « Reply #2 on: Apr 10th, 2003, 2:04am » Quote Modify

I know this one is on the site somewhere...
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ThudnBlunder
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 Re: Rope Around the Earth   « Reply #3 on: Apr 10th, 2003, 2:48am » Quote Modify

Quote:
 I know this one is on the site somewhere...

You are probably mistaking this for a similar (but much easier) puzzle.
 « Last Edit: Apr 10th, 2003, 9:58am by ThudnBlunder » IP Logged

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wowbagger
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 Re: Rope Around the Earth   « Reply #4 on: Apr 10th, 2003, 4:37am » Quote Modify

on Apr 9th, 2003, 8:12pm, Icarus wrote:
 For a 40,000 Km circumference of the Earth, I get [hidden answer]

This is impossible. I'm sure you mean another riddle.
Consider the fact that the increase in length has to be at least twice the height you reach - but we only added one metre.

My result with r = 6370km is about 78.5 10-6m, but I'm not too sure about it.
 « Last Edit: Apr 10th, 2003, 4:38am by wowbagger » IP Logged

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aero_guy
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 Re: Rope Around the Earth   « Reply #5 on: Apr 10th, 2003, 4:52am » Quote Modify

Nope, they're right.  Consider a rope going across a very long flat plane.  If you add something to the length and then pull it up at the middle you get two identical triangles, each with base b, height you pulled them up a, and hypotenuse c.  If the rope was originally 200000m and you added 2m, then c=100001m

a=sqrt(c2-b2)

so you see, by increasing c by 1m, we have increased c2 by 200001m, thus making a, the height obove the plane we pull the rope to, rather large.

I think you missed that when you stretch the rope, it will leave your hand to intersect the edge of the world at a tangent.
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wowbagger
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 Re: Rope Around the Earth   « Reply #6 on: Apr 10th, 2003, 5:07am » Quote Modify

on Apr 10th, 2003, 4:52am, aero_guy wrote:
 Nope, they're right.  Consider a rope going across a very long flat plane.  If you add something to the length and then pull it up at the middle you get two identical triangles, each with base b, height you pulled them up a, and hypotenuse c.  If the rope was originally 200000m and you added 2m, then c=100001m   a=sqrt(c2-b2)   so you see, by increasing c by 1m, we have increased c2 by 200001m, thus making a, the height obove the plane we pull the rope to, rather large.

I understand your calculation. If your rope in the plane were the size of the earth's circumference, the height you can reach by adding 1m of rope would be ca. 141m.
So the seemingly easy way to reject Icarus's result doesn't work.

Quote:
 I think you missed that when you stretch the rope, it will leave your hand to intersect the edge of the world at a tangent.

I'm not sure whether I understand this part.
Of course I considered the fact that the rope runs around the earth for a rather large angle before leaving the surface at some points A and B. The lines from A and B to the point P where you hold the rope with your hand are tangents to the earth. From there I used some geometry and algebra, but - as I said - I'm not sure about my result.
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 Re: Rope Around the Earth   « Reply #7 on: Apr 10th, 2003, 5:30am » Quote Modify

on Apr 10th, 2003, 4:37am, wowbagger wrote:
 Consider the fact that the increase in length has to be at least twice the height you reach - but we only added one metre.

This is wrong. Even I have noticed this by now. I mistook an upper bound for a lower (or vice versa).
Quote:
 My result with r = 6370km is about [hidden try]

This is also wrong. Even using my formula and my value of r, I now calculate a different (still smaller) value.
 « Last Edit: Apr 10th, 2003, 5:31am by wowbagger » IP Logged

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towr
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 Re: Rope Around the Earth   « Reply #8 on: Apr 10th, 2003, 6:50am » Quote Modify

at the moment I'm getting  121.4m (I used the wrong diameter before) [/edit]
But I'm still not sure I'm close enough with my approximation..

Here's what I've done:

the part of the rope that touches the sphere + the part of rope that's in the air equals the diameter of the sphere + one
this gives:
(2pi - 2a)r + 2 tan(a)r = 2*pi*r +1
you can then go use
an+1 = arctan((2r*an+1)/2r)

and then r(1/(cos(a) - 1) provides the answer
 « Last Edit: Apr 10th, 2003, 7:21am by towr » IP Logged

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aero_guy
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 Re: Rope Around the Earth   « Reply #9 on: Apr 10th, 2003, 8:34am » Quote Modify

My equations come out the same, but solving them gives me an a of .003347 which gives a height of 224m
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towr
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 Re: Rope Around the Earth   « Reply #10 on: Apr 10th, 2003, 10:52am » Quote Modify

Have you tried approximating it from both sides? You may not have converged to a stable value yet.
Though it's of course possible there's a mistake in my value (though I'd expect matlab to do rather well).

Also, you could do the reverse calculations, and check if you get back to 40000001 meters.. (which is the case for my value. But again, it could be freak rounding errors..possibly..)
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 Re: Rope Around the Earth   « Reply #11 on: Apr 10th, 2003, 11:05am » Quote Modify

I guess it is possible that the rounding errors are reproducing themselves when working backwords, but on the way back I still get .99953m, close enough for me.
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towr
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 Re: Rope Around the Earth   « Reply #12 on: Apr 10th, 2003, 11:34am » Quote Modify

peculiar..
Are we both using an r of 40 000 000/(2*pi)? [e]removed extra 0[/e]
And which program do you use to calculate it?

 Ah, i see, I think you used r = 40 000 000..
That's actually the same mistake I made the first time (but I had forgotten that value allready)..

On another try I got Icarus' answer of 192, but I made a mistake there as well.. I used an+1 = arctan((r*an+1)/r) by mistake that time..

I'm sure r actually has to be as above..
[/edit]
 « Last Edit: Apr 10th, 2003, 11:14pm by towr » IP Logged

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aero_guy
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 Re: Rope Around the Earth   « Reply #13 on: Apr 10th, 2003, 11:47am » Quote Modify

my mistake, I was using the number as r.  That would be a very large earth.  I get 121.4m now.
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 Re: Rope Around the Earth   « Reply #14 on: Apr 10th, 2003, 1:36pm » Quote Modify

I, too, am getting 121.43m (using matlab).

Honestly, I don't think it belongs in the "hard" section... even if I (or others) made an arithmetic mistake.
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Icarus
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 Re: Rope Around the Earth   « Reply #15 on: Apr 10th, 2003, 5:11pm » Quote Modify

on Apr 10th, 2003, 11:34am, towr wrote:
 peculiar.. Are we both using an r of 400000000/(2*pi)? And which program do you use to calculate it?    Ah, i see, I think you used 40 000 000.. That's actually the same mistake I made the first time (but I had forgotten that value allready)..   I'm sure r actually has to be as above.. [/edit]

The Earth is only 40,000 Km in circumference, not 400,000 Km.
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 Re: Rope Around the Earth   « Reply #16 on: Apr 10th, 2003, 7:53pm » Quote Modify

Yes, BNC, despite the number of errors people are making, this does not belong in the hard section.  There are many questions in the Easy section tougher than this. William Wu's main page says these are "hard core" riddles, so easy doesn't mean trivial. In the "general problem solving/whatever" forum there is a thread about what people mean by easy.

I also come up with 121.43 m. The error in Icarus' solution seems to be that he used the full meter of rope to be in one triangle, instead of dividing it between the two triangles.  Figuring out what Icaraus did wrong when he didn't show his work-- that was not so easy.

The only tricky thing about this riddle is solving an equation of the form x+a=tan(x) for x. However that can be done by trial and error, with special software, or iteratively like towr did.  By expanding tan(x) into a series, some pretty good approximations (good for small a, which it is in this problem) can be found:  x=(3a)1/3 even better is

The height the rope lifts off earth is

where r is radius of earth, a=A/2/r, A is the additional length of rope (1 meter in this case).  The first approximation leads to 121.431 m, and the second gives 121.430.  The second approximation results in a value of x that makes tan(x) within 1e-17 of x+a, so it is fairly accurate for this small value of a.
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towr
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 Re: Rope Around the Earth   « Reply #17 on: Apr 10th, 2003, 11:08pm » Quote Modify

on Apr 10th, 2003, 5:11pm, Icarus wrote:
 The Earth is only 40,000 Km in circumference, not 400,000 Km.
So I typed an extra 0 in my post, if you check the numbers you'll see that wasn't the case in my calculations :p
And you'll notice  I did type it right the second time..

It was the factor 2*pi that was at issue btw, not the 40 something something kilometers..
 « Last Edit: Apr 10th, 2003, 11:09pm by towr » IP Logged

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 Re: Rope Around the Earth   « Reply #18 on: Apr 11th, 2003, 3:46pm » Quote Modify

towr - Sorry I didn't read your post carefully enough. I thought you were saying 40,000,000 was the mistake and should have been 400,000,000. I see now what you meant.
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NickH
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 Re: Rope Around the Earth   « Reply #19 on: Apr 22nd, 2003, 12:38pm » Quote Modify

I agree this puzzle doesn't belong in the hard section.

See the solution on my site, where I show that the height reached is approximately:

k · r1/3 · d2/3, where k = (3/2)2/3/2, r is the Earth's radius, and d is the extra rope length.
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