A question about the particular solution to the Catenary problem

Question

The general solution to Catenary is $$y = \frac{1}{a} \cosh\,a(x+C_1)+C_2$$ where a, C1 and C2 are unknown parameters. Now I have three initial conditions: $(0,0)$, $(x_0,y_0)$ and $$\sqrt{22.05^2-y_0^2}=\frac{2}{a}\sinh\frac{ax_0}{2}$$ how can I figure out the particular values of $a, C_1$ and $C_2$? Thanks a lot.

Answer 1

You have to solve the system of 3 equations for $a$ , $C_1$ , $C_2$ : $$\begin{cases} 0 = \frac{1}{a} \cosh(aC_1)+C_2 \\ y_0 = \frac{1}{a} \cosh(ax_0+aC_1)+C_2\\ \sqrt{c^2-y_0^2}=\frac{2}{a}\sinh\frac{ax_0}{2}\quad\text{with } c=22.05 \end{cases}$$ $C_2= -\frac{1}{a}\cosh(aC_1) \quad\to\quad \begin{cases} ay_0 = \cosh(ax_0+aC_1)-\cosh(aC_1)\\ \sqrt{c^2-y_0^2}=\frac{2}{a}\sinh\frac{ax_0}{2} \end{cases}$

In the equation $\quad \sqrt{c^2-y_0^2}=\frac{2}{a}\sinh\frac{ax_0}{2}\quad$ all parameters are known, except $a$. So, it is a non-linear equation with one unknown.

Let $X=\frac{ax_0}{2} \quad\to\quad \frac{\sinh(X)}{X} =Y_0=x_0\sqrt{c^2-y_0^2}$

$Y_0$ is known, which allows to compute $X$:

On analytical viewpoint, the inverse fonction of $\frac{\sinh(X)}{X}$ being denoted $\text{sinhc}$, then $X=\text{sinhc}(Y_0)$ $$a=\frac{2}{x_0}\text{sinhc}(Y_0)=\frac{2}{x_0}\text{sinhc}\left(x_0\sqrt{c^2-y_0^2}\right)$$ On practical viewpoint, of course we don't use this symbolism, but we solve the equation thanks to numerical calculus.

Then, bringing back the known value of $a$ into : $$ay_0 = \cosh(ax_0+aC_1)-\cosh(aC_1)$$ all parameters are known, except $C_1$. So, it's a non-linear equation with one unknown. Again, we have to solve it for $C_1$ thanks to numerical calculus.

Then, with the known values of $a$ and $c_1$, we compute : $$C_2= -\frac{1}{a}\cosh(aC_1)$$