Csdrhrt

I need to find minimum point of hanging rope with two known points $p_1, p_2$ (start and end point of the rope) and known rope length.

I want to model all rope shapes with different length and start and end point.

Do I have to use numerical methods?

Does it have any closed form solution?

Known $to L,(x_1,y_1), (x_2,y_2)$ then $a=$?

I use a general catenary equation like below:

$$
f(x)= acoshleft(frac {x-b}aright)+c\
L= asinhleft(frac{x_2}{a}right)-asinhleft(frac{x_1}{a}right)
$$

Calling

$$
x_2-x_1 = h\
y_2-y_1 = v
$$

we have

$$
v = acoshleft(frac{x_2}{a}right)-acoshleft(frac{x_1}{a}right)\
L = asinhleft(frac{x_2}{a}right)-asinhleft(frac{x_1}{a}right)
$$

so after some trigonometric transformations we arrive at

$$
L^2-v^2 = 4a^2sinh^2left(frac{x_2-x_1}{2a}right)
$$

so we have finally

$$
L^2-v^2 = 4a^2sinh^2left(frac{h}{2a}right)
$$

with this last equation we can determine the $a = a_0$ value using an iterative procedure.

NOTE

Calling now

$$
y(x) = y_0 + a_0coshleft(frac{x-x_0}{a_0}right)
$$

such that

$$
y(x_1) = y_1\
y(x_2) = y_2
$$

The minimum point is obtained by solving for $x$

$$
y'(x) = 0
$$

so if $x_1 le x_0 le x_2$ follows

$$
min y(x) = y_0 + a_0
$$

otherwise

$$
min y(x) = min(y_1, y_2)
$$

Now, regarding the iterative process we have

$$
frac{L^2-v^2}{4a^2} = sinh^2left(frac{h}{2a}right)
$$

or

$$
C_0^2lambda^2=sinh^2lambda, lambda = frac{h}{2a}, C_0 = frac{sqrt{L^2-v^2}}{h}
$$

or

$$
C_0 lambda = sinh(lambda)
$$

which should be solved numerically.

Starting from @Cesareo answer, considering
$$L^2-v^2 = 4a^2sinh^2left(frac{h}{2a}right)$$ let $x=frac{h}{2a}$ to make the equation
$$frac{L^2-v^2}{h^2}=frac{sinh^2(x) }{x^2}implies k=sqrt{frac{L^2-v^2}{h^2}}=frac{sinh(x) }{x}$$

Consider that you look for the zero of $$f(x)=frac{sinh(x) }{x}-k$$ It varies very quickly. Then, it would be better to look for the zero of
$$g(x)=log left(frac{sinh (x)}{x}right)-log(k)$$ which is much better conditioned.

Edit

To get a starting value $x_0$ for Newton method, using Padé approximant built at $x=0$
$$frac{sinh (x)}xsimeq frac{7 x^2+60}{60-3 x^2} implies x_0=frac{2 sqrt{15} sqrt{k-1}}{sqrt{3 k+7}}$$ which would be good for $1 leq k leq 3$.

For larger values of $k$
$$frac{sinh (x)}xsimeq frac{e^x}{2 x}implies x_0=-W_{-1}left(-frac{1}{2 k}right)$$ where appears Lambert function.

The table below shows some results
$$left(
begin{array}{ccc}
k & x_0 & x \
1.25 & 1.18125 & 1.18273 \
1.50 & 1.61515 & 1.62213 \
1.75 & 1.91663 & 1.93300 \
2.00 & 2.14834 & 2.17732 \
2.25 & 2.33550 & 2.37963 \
2.50 & 2.49136 & 2.55265 \
2.75 & 2.62398 & 2.70395 \
3.00 & 2.73861 & 2.83845 \
& & \
3.00 & 2.83315 & 2.83845 \
3.25 & 2.95545 & 2.95952 \
3.50 & 3.06642 & 3.06962 \
3.75 & 3.16801 & 3.17058 \
4.00 & 3.26169 & 3.26380 \
4.25 & 3.34861 & 3.35037 \
4.50 & 3.42970 & 3.43117 \
4.75 & 3.50567 & 3.50693 \
5.00 & 3.57715 & 3.57823
end{array}
right)$$

Edit

Looking at this question (which I did not remember - problem of age, I guess), I noticed that I was able to generate a quite good estimates building at $x=0$ the $[3,4]$ Padé approximant of $sinh(x)-k x$.

From this, I considered building the $[3,2n]$ Padé approximants which write
$$sinh(x)-k x=x frac{(1-k)+a^{(n)}_1 x^2 }{1+sum_{m=1}^n b_m x^{2m} }$$ leading to an approximate solution
$$x=sqrt{frac {k-1}{a^{(n)}_1 }}$$ For sure, this was done using a CAS. The longest result able to fit on a single line corresponds to $n=6$ and the result is
$$x=frac{sqrt{6} sqrt{(k-1)(105 k^5+60705 k^4+1365738 k^3+5507466 k^2+5665509 k+1414477 )}}{sqrt{3 k^6+6120 k^5+307017 k^4+2586544 k^3+5952621 k^2+4301640
k+860055}}$$ which seems to be very good even for large values of $k$ (checked up to $k=500$).