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Say I have a connected at either end to two points, $A(x_A, y_A)$ and $B(x_B, y_B)$ of length $l$, where $l \leq \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$, how would I go about finding an equation of its shape?

I guess the chain would be of the form $y = \alpha \cosh{(ax + b)} + \beta$, so how do the coefficients relate to the coordinates of the hanging points and its length?

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The general shape is

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

where $(x_c,y_c)$ is the lowest point on the curve (sag point) and $a = \frac{H}{w}$ is the catenary constant. Here $H$ is the horizontal tension shared along the cable, and $w$ is the unit weight (that is weight over length).

Consider a span $S$ and height $h$ the above is fit to the boundary conditions $y(0)=0$ and $y(S)=h$ with

$$ x_c = \frac{S}{2} - a\, \sinh^{-1} \left( \frac{h\,\exp(S/2a)}{a\, \exp(S/a)-a} \right) $$ $$ y_c = a \left( \cosh\left(\frac{x_c}{a}\right) -1 \right) $$

Furthermore, the tension in the cable is defined by the constant horizontal component $H$, and the varying vertical component $$V(x) = H \sinh\left( \frac{x-x_c}{a} \right) $$ as well as the total tension $$T(x) = \sqrt{H^2+V^2}= H \cosh\left( \frac{x-x_c}{a} \right) $$

To find the tension at the ends, use $x=0$ and $x=S$. If the vertical tension at the ends is negative then there is an uplift condition where the cable is trying to pull the support out of the ground.

Finally, the total length is

$$ L = \int {\rm d}s = \int \limits_0^S \sqrt{1+y'(x)^2}\,{\rm d}x = a \left( \sinh\left( \frac{S-x_c}{a} \right) + \sinh\left( \frac{x_c}{a} \right) \right) $$


References:

Example software catenary solver based on the above equations enter image description here

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Taking the solution you propose, you can make $b=0$ since

$cosh(ax+b)=\dfrac{e^{ax+b}+e^{-ax-b}}{2}$

and you can absorb $\beta$ to the remnant of the expansion for the expression. An analogous can be done for constant $\alpha$, and set $\alpha=1$ (Think out: This problem is about two forces acting in one plane, the weight of the chain and the tension of one extreme attached to the ceiling, so it will result on a 2nd order differential equation which will only need 2 constants to determine! as for it is an initial value problem).

Thus, we only need to determine $a$ and $\beta$.

From then, $y(0)$ would be the height the chain has when fully extended and both forces (the tension of one extreme and the weight of the chain) act upon it. From then, $y(0)=\beta$. If the chain is uniform, you can get that in equilibrium:

$l\,T=\beta\,W$

where $T$ would be the tension and $W$ the weight, both per unit length of the chain. From this you get constant $\beta$.

Finally, you can get $a$ when, for example, you set $\dfrac{dy}{dx}=0$ and the chain remains still when hanging.

Also if you are keen on Variational Calculus you may find the way to compute it, you can check chapter 6 from Thornton-Marion's "Classical Dynamics of Particles and Systems" for a good introduction towards it and the same problem.

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