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Say you're standing on the equator and you have a string below you tied around the equator (40,075 km) that is the length of the equator + 1 meter (40,075.001 km). What is the maximum height you can you lift the string off the ground? Can you create a function of both circumference of the circle (earth) and string to output the distance between the two if pulled tight?

Assumptions:

  • For illustration, the result would be pulled from a single point, making a triangle until it met with the earth, in which it would follow the curvature of the earth. Similar to a snow-cone or O>
  • The string does not stretch
  • The earth can be assumed to be a perfect sphere
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  • $\begingroup$ How tight is the string? $\endgroup$ – user328032 Apr 18 '16 at 18:50
  • $\begingroup$ Do you know the formula of the circumference of a circle ? $\endgroup$ – callculus Apr 18 '16 at 18:55
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    $\begingroup$ We get a transcendental equation. One can get an excellent approximation to the solution by using a power series expansion. $\endgroup$ – André Nicolas Apr 18 '16 at 19:05
  • $\begingroup$ I'm far from a math major, and it took me about a day to solve, but you're correct in that the best I could do was a very good approximation. $\endgroup$ – Travis Apr 18 '16 at 19:09
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    $\begingroup$ See also mathforum.org/mathimages/index.php/Rope_around_the_Earth. $\endgroup$ – lhf Apr 19 '16 at 0:19
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We have to solve (with respect to $d$) the trascendental equation: $$ 2\sqrt{(R+d)^2-R^2}+2R\arcsin\frac{R}{R+d}=\pi R+1 $$ or solve (with respect to $x=\frac{d}{R}$) the trascendental equation: $$ 2\sqrt{x^2+2x}+2\arcsin\frac{1}{1+x}=\pi+\frac{1}{R} $$ that is equivalent to: $$ \sqrt{x^2+2x}-\arctan\sqrt{x^2+2x}=\frac{1}{2R}.$$ In order to solve $z-\arctan z=\frac{1}{2R}$ we may apply Newton's method, then $x=-1+\sqrt{z^2+1}$.

Since $z$ is close to zero, $z-\arctan z\approx\frac{z^3}{3}$, hence: $$ d\approx \sqrt[3]{\frac{9R}{32}}\approx \color{red}{121.5\,m}.$$

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Draw a picture. Let $P$ be the peak of the stretched string, and let $C$ be the centre of the Earth. Let $G$ be the point on the Earth's surface which is on the line $PC$, and let $T$ and $T'$ be the points of tangency of the string with the Equator after the string has been pushed upwards to full height $PG$.

Let $r$ be the radius of the Earth, and let $\epsilon r$ be the excess amount of string in addition to $2\pi r$. Note that $\epsilon$, in your example, is very small.

Let $\theta=\angle TCG$ Note that $\theta$ is small. The key equation is $$\epsilon r=2r\tan\theta -2r\theta.$$ This holds because $2r\tan\theta$ is the sum of the lengths of the two tangent segments $PT$ and $PT'$, while $2 r\theta$ is the amount of string saved because it no longer covers the minor arc $TGT'$. The difference is equal to the amount $\epsilon r$ of extra string we have available.

Using the fact that $\tan\theta\approx \theta +\frac{\theta^3}{3}$, we get $$\theta\approx \left(\frac{3\epsilon}{2}\right)^{1/3}.$$

Now that we know $\theta$, we can find the height $PG$ of the peak of the string. From the diagram, we can see that $PG=r(\sec\theta-1)$. Since $\theta$ is small, this is well approximated by $\frac{\theta^2}{2}$, and we end up with $$PG\approx \frac{r}{2}\left(\frac{3\epsilon}{2}\right)^{2/3}.$$

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  • $\begingroup$ We end up with $$PG\approx \frac{r}{2}\left(\frac{3\delta}{2r}\right)^{2/3}?$$ $\endgroup$ – Narasimham Apr 26 '16 at 21:15
  • $\begingroup$ @Narasimham: Note that I defined $\delta r$ to be the excess amount of string. If you let $\Delta$ be the excess amount, then $\Delta=\delta r$, and we get your expression, with $\frac{3\Delta}{2r}$ instead of your $\frac{3\delta}{2r}$. The reason I defined $\delta$ the way I did was "dimensional." I wanted to end up with an expression that was a constant times $r$. I will check later whether there is a mistake, but dimensionally the formula I gave has the right shape. $\endgroup$ – André Nicolas Apr 26 '16 at 21:22
  • $\begingroup$ Right, my bad. I would use a symbol like $\epsilon$ Shall delete my remark. $\endgroup$ – Narasimham Apr 27 '16 at 3:05
  • $\begingroup$ @Narasimham: I had hesitated between using $\delta$ and using $\epsilon$! Will take that as a suggestion and make the change. $\endgroup$ – André Nicolas Apr 27 '16 at 3:14
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EDIT 1:

Let $\Delta C$ be the extra length of loose string.

$$ r = \frac{C}{2 \pi} ;\ \Delta r = \frac{\Delta C}{2 \pi} = h= \frac{1000\, mm}{2 \pi} = 159. 155 \; {mm \,!} $$

The relation/ratio is the same on Mars or Saturn or any earth based large round cylinder.

In this problem you want $ h $ for given input $ (R, \Delta C) $

We calculate tangent length $T$ shallow horizon range using right triangle property :

$$ T^2 = h ( 2 R -h ) \tag{1} $$

$$ T = R \tan \theta \tag{2}$$

And eliminate $T$ to solve for $ \tan \theta $

$$ \tan \theta= \sqrt{ (h/a) (2-h/a) } \tag{3}$$

From the diagram the slack above involute is indicated by red line,

$$ = IE =\tan \theta - \theta = \Delta C/ (2 h) $$

or,

$$ \sqrt { (h/a) (2-h/a) } -\tan^{-1}{\sqrt{ (h/a) (2-h/a) } } = \Delta C/(2 R) \tag {4} $$

Plugging in numerical values to solve for $h$

$$ R = 40075 * 10^3/( 2 \pi) \, meters ; \, \Delta C =1 m \rightarrow h = 121.508 m \tag{5}$$

$\because h << R$ an approximation using Madhava-Gregory-Leibnitz series

$$ \tan u - \tan ^{-1}u \approx u^3/3 $$

gives us $$ h^3 =\frac {9}{32}\cdot {\Delta C}^2 \cdot R \tag {6}$$

giving only a bit different value

$ h= 121.505 m $.

Involute+Slack

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Half a meter. You take the slack extra meter and lift it half a meter high.

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