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My question is the following:

What is the shape of the rope which holds a kite flying? (Steady state.)

I am not a physicist (I am a mathematician), so I can not work the physics part of the question.

My guess is that it should be the same as catenary (hyperbolic cosine)

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  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/51485/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Mar 3, 2014 at 20:52
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    $\begingroup$ Your guess is correct $\endgroup$
    – pho
    Commented Mar 3, 2014 at 20:56
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    $\begingroup$ ""it should be the same as catenary (hyperbolic cosine) "" It is not the same, it is a catenary, but only in math. In physics You have to know the drag of the wind on every piece of the rope. Because this drag is a rather nonlinear function of wind velocity, things are complicated. $\endgroup$
    – Georg
    Commented Mar 3, 2014 at 20:58
  • $\begingroup$ @Georg but you know that in physics, all ropes are massless, frictionless, 1-dimensional objects. :-) . Tho' in this case I guess we have to give it mass (so the catenary doesn't collapse to a line segment), but let's keep it frictionless and inextensible. $\endgroup$
    – Carl Witthoft
    Commented Mar 3, 2014 at 21:05

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The catenary shape between two points with coordinates $A=(0,0)$ and $B=(S,h)$ is

$$ y(x) = y_C + a \left( \cosh \left( \frac{x-x_C}{a} \right)-1 \right) $$

where

$$ \begin{gather} a = \frac{H}{w} & \mbox{Catenary Constant} \\ H : & \mbox{Horizontal Tension} \\ w : & \mbox{Weight per Unit Length} \\ x_C = \frac{S}{2} + a \sinh \left( \frac{h \exp(\eta)}{a (1-\exp(2\eta))} \right) & \mbox{Lowest Point x coordinate} \\ y_C = -a \left( \cosh\left( -\frac{x_C}{a}\right)-1\right) & \mbox{Lowest Point y coordinate} \\ \eta = \frac{S}{2 a} & \mbox{strain factor} \end{gather} $$

The results look like this: Catenary

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  • $\begingroup$ I've been (idly) wondering about an extension of these solutions: what is the value of $h$ ( the relative height of the second anchor point) at which the catenary's lowest point equals $h$ -- now all I have to do is find the extrema of $y_C$ $\endgroup$
    – Carl Witthoft
    Commented Feb 28, 2016 at 14:39
  • $\begingroup$ Yes, that is the tower uplift condition and it occurs when $x_c=0$ or $x_c=S$ depending on the side it happens. You can solve for it and find the constant $a$ that this happens for (numerically), or $h$ itself. The solution I got was $$ h = \pm a \left(\cosh\left(\frac{S}{a}\right)-1\right) $$ $\endgroup$
    – John Alexiou
    Commented Feb 28, 2016 at 22:10

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