Puzzle for you

[Albert Chan]

(08-17-2018 03:59 AM)Thomas Klemm Wrote:  We can model the rope as a catenary:

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

...

Example:

\(h=3\)

\(s=4\)

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

Out of curiosity, had the curve a parabola, with same h and x, s ~ 3.409 (less rope)

With a fixed h and x, can I assume catenary always curvier than a parabola ?

Regarding this puzzle, my original guess were also 0 distance between poles.
But, the rope cannot even fit between the poles ...

Edit: catenary is indeed curvier (flatter bottom, steeper rise): http://mathforum.org/library/drmath/view/65729.html