Puzzle for you

[Thomas Klemm]

We can model the rope as a catenary:

\(y=a\cosh(\frac{x}{a})\)

where \(a\) is a scaling factor.

Then with \(h\) as the difference of the heights and \(s\) as half the length of the rope we get:

\(h=a(\cosh(\frac{x}{a})-1)\)

\(s=a\sinh(\frac{x}{a})\)

With:

\(u=\frac{x}{a}\)

and

\(v=\frac{h}{s}=\frac{a(\cosh(u)-1)}{a\sinh(u)}=\tanh(\frac{u}{2})\)

Thus

\(u=2\tanh^{-1}(v)=2\tanh^{-1}(\frac{h}{s})\)

This allows us to calculate:

\(a=\frac{s}{\sinh(u)}=\frac{s}{\sinh(2\tanh^{-1}(\frac{h}{s}))}=\frac{s^2-h^2}{2h}\)

Now we plug both \(a\) and \(u\) in to calculate:

\(\begin{align*}
x=a\cdot u &=\frac{s^2-h^2}{2h}\cdot2\tanh^{-1}(\frac{h}{s}) \\
&=\frac{s^2-h^2}{h}\cdot\tanh^{-1}(\frac{h}{s}) \\
&=s\cdot(\frac{s}{h}-\frac{h}{s})\cdot\tanh^{-1}(\frac{h}{s}) \\
&=s\cdot(\frac{1}{v}-v)\cdot\tanh^{-1}(v)
\end{align*}\)

Example:

\(h=3\)

\(s=4\)

\(x=4\cdot(\frac{4}{3}-\frac{3}{4})\cdot\tanh^{-1}(\frac{3}{4})\approx 2.27022850723\)

Limit \(h\to s\)

For \(v\to1\) the value \(\tanh^{-1}(v)\to\infty\) but at the same time \(\frac{1}{v}-v\to0\).
Thus we can apply L'Hôpital's rule to find:

\(\begin{align*}
\lim_{v\to1}(\frac{1}{v}-v)\cdot\tanh^{-1}(v) &= \lim_{v\to1}\frac{1-v^2}{v}\cdot\tanh^{-1}(v) \\
&= \lim_{v\to1}\frac{\tanh^{-1}(v)}{\frac{v}{1-v^2}} \\
&= \lim_{v\to1}\frac{\frac{1}{1-v^2}}{\frac{1+v^2}{(1-v^2)^2}} \\
&= \lim_{v\to1}\frac{1-v^2}{1+v^2}=\frac{1-1}{1+1}=0
\end{align*}\)

And thus:

\(x=s\cdot0=0\)

Which we already knew.

Cheers
Thomas