Second end of a Catenary of known length passing through a point and intersecting two straight lines at a known distance.

Question

I have the coordinates of one extreme point $D=(0,y_D)$, two lines $y = m_ix + y_D$ for $i\in \{ A,B \}$ intersecting the catenary at two point $A$ and $B$, and also in $D$. I know the length $L_{DA}$ of the catenary from $D$ to $A$, $L_{AB}$ from $A$ to $B$ and the overall length $L_{DE}$ from $D$ to $E$. I want to find the coordinates of the second extreme point $E$, as per the linked picture ($L = L_{AB}$).

My understanding is that given these conditions (one point, two intersections and a known distance between the intersections) the solution must be unique. My approach would be to find the equation of the catenary and then locate $E$, $L_{DE}$ away from the starting point $D$ along the curve.

I know that there is probably no analytical solution, as this involves a transcendental function. The equations I derived read as follows:

The shape of the catenary is expressed as $$y = c + a \cosh\left(\frac{x-b}{a}\right),$$

where $a,b,c$ are unknown. Three lengths are known

$$ L_{AB} = \int_{x_A}^{x_B} \cosh\left(\frac{t-b}{a}\right)dt = a\left[\sinh\left(\frac{x_B -b}{a}\right) - \sinh\left(\frac{x_A -b}{a}\right)\right] $$ $$ L_{DA} = \ldots = a\left[\sinh\left(\frac{x_A -b}{a}\right) - \sinh\left(\frac{-b}{a}\right)\right],$$ $$ L_{DE} = \ldots = a\left[\sinh\left(\frac{x_E -b}{a}\right) - \sinh\left(\frac{-b}{a}\right)\right],$$ and the two lines intersect the catenary $$y_A = c + a \cosh\left(\frac{x_A-b}{a}\right) = m_A x_A + y_D$$ $$y_B = c + a \cosh\left(\frac{x_B-b}{a}\right) = m_B x_B + y_D$$

I have looked at similar problems, such as Finding points along a catenary curve, but I do not know how to reduce this problem to that one. It seems to me the two coincide, once the coordinates $(x_A,y_A)$ or $(x_B, y_B)$ have been fixed. Is it possible to obtain a similar solution (i.e. analytically driven, but for the solution of the transcendental equation) or an algorithm to approximate the coordinates of $E$?

Answer 1

I think that the problem can be solved as a single optimization problem using as only variables $a$ and $b$

The coordinates of the points are $$x_A \qquad\qquad y_A=y_D+a \cosh \left(\frac{x_A-b}{a}\right)-a \cosh \left(\frac{b}{a}\right)$$ $$x_B \qquad\qquad y_B=y_D+a \cosh \left(\frac{x_B-b}{a}\right)-a \cosh \left(\frac{b}{a}\right)$$ $$x_E \qquad\qquad y_E=y_D+a \cosh \left(\frac{x_E-b}{a}\right)-a \cosh \left(\frac{b}{a}\right)$$

We also have $$y_A=y_D+m_A\,x_A \qquad\text{and}\qquad y_B=y_D+m_B\, x_B$$ So, the five equations are $$\text{Equa}_1=a \cosh \left(\frac{x_A-b}{a}\right)-a \cosh \left(\frac{b}{a}\right)-m_A \,x_A=0\tag 1$$ $$\text{Equa}_2=a \cosh \left(\frac{x_B-b}{a}\right)-a \cosh \left(\frac{b}{a}\right)-m_B \,x_B=0\tag 2$$ $$\text{Equa}_3=a\sinh\left(\frac{x_B -b}{a}\right) - a\sinh\left(\frac{x_A -b}{a}\right)-L_{AB}=0\tag 3$$ $$\text{Equa}_4=a\sinh\left(\frac{x_A -b}{a}\right)+ a\sinh\left(\frac{b}{a}\right)-L_{DA}=0\tag 4$$ $$\text{Equa}_5=a\sinh\left(\frac{x_E -b}{a}\right)+ a\sinh\left(\frac{b}{a}\right)-L_{DE}=0\tag 5$$

Using $(4)$ and $(5)$ we have $$x_A=b+a \sinh ^{-1}\left(\frac{L_{DA}}{a}-\sinh \left(\frac{b}{a}\right)\right)\tag 6$$ $$x_E=b+a \sinh ^{-1}\left(\frac{L_{DE}}{a}-\sinh \left(\frac{b}{a}\right)\right)\tag 7$$

Using $(3)$ $$x_B=b+a \sinh ^{-1}\left(\sinh \left(\frac{x_A-b}{a}\right)+\frac{L_{AB}}{a}\right)\tag 8$$

So $(x_A,x_B,x_E)$ are explicit functions of $(a,b,L_{AB},L_{DA},L_{DE})$

So, consider the norm $$\Phi(a,b)=\big[\text{Equa}_1\big]^2+\big[\text{Equa}_2\big]^2$$ and minimize it with respect to $a$ and $b$.

Edit

I made a test run using $$m_A=-\frac 12 \quad m_B=-\frac 3{10}\quad L_{AB}=\frac{5773}{10000}\quad L_{DA}=\frac{34621}{10000}\quad L_{DE}=\frac{55743}{5000}$$ and the results are $$a=1.199971086\qquad b=1.799969533\qquad \implies \qquad \Phi \sim 0$$ which leads to $x_E=4.999926081$.

Update

As I wrote in comments, it would be better to consider as independant variables $a$ and $k=\frac ba$. Doing so, we can divide $(1)$ and $(2)$ by $a$ and have $$m_A\Bigg[k+\sinh ^{-1}\left(\frac{L_{DA}}{a}-\sinh (k)\right) \Bigg]+\cosh(k)-\sqrt{1+\left(\frac{L_{DA}}{a}-\sinh (k)\right)^2}=0 \tag {1a}$$ and a similar one to make $(2a)$

Answer 2

I found at least a numerical solution to the problem. I use a nonlinear, constrained optimizer to minimize an error cost function of the parameters $a$, $b$. Given that the catenary must go through $(0, y_D)$, parameter $c$ is a function of $a$ and $b$: $$ c = y_D - a \cosh\left(\frac{-b}{a}\right).$$

The error cost function $e$ is nonlinear, and is the sum of the norms of the errors between the known and estimated distances of $D$ and $A$ and of $A$ and $B$ along the catenary. Using the equations in the question text and recalling that $x_D=0$ by definition:

$$e = || a\left(\sinh\left(\frac{x_B-b}{a}\right) - \sinh\left(\frac{x_A-b}{A}\right)\right) - L_{AB}|| + ||a\left(\sinh\left(\frac{x_A-b}{a}\right) - \sinh\left(\frac{-b}{A}\right)\right) - L_{DA}||.$$

At each call of the cost function, with a new pair of parameters $a$ and $b$, $x_A$ and $x_B$ are calculated numerically by solving $$c + a \cosh\left(\frac{x_A-b}{a}\right) = m_A x_A + y_D$$ with Newton's method, as suggested here. Upper and lower bounds for optimization variables $a$ and $b$ depend on the application. Furthermore, I suggest to pick the initial guess $a_0, b_0$ as an underestimate of what the values are going to be. In my experience, the optimization problem tends to converge to the correct solution more easily this way. The cost function is highly non-linear, therefore convergence depends (also) on the initial guess.

Answer 3

If I did not miss something, the following:

Arc lengths (DA,DB,DE) measured from D are given. Coordinates of (A,B,E) should be found. Lines connecting D to them may be drawn as a last (avoidable) step.

If we find the differential equation of all Catenaries passing through D we notice that there is no unique determination of coordinates.. but can be found upto constants of integration.

The Catenary ode can be derived intrinsic:

Catenary property $ y \cos \phi = c$ is differentiated w.r.t. arc and simplified.

$$ \frac{d \phi}{ds}= \frac{\cos^{2}\phi}{c}\; ;$$ Integrating

$$ \tan \phi = \frac{s-s_{min}}{c} \; ; $$

in terms of slope $\phi$, arc length s, arc-length $s_{min}$ at minimum ordinate c. Else this is also obtained by differentiating

$$ y= c \cosh \frac{x-C_2}{c}. $$

There would be as many solutions as the number of assumptions for admissible initial slopes $\phi_{D}.$ If $c$ is also varied then there is a doubly infinite set of solutions.

In other words, each Catenary can be bent to any other of the Catenary set passing through D. Some cases can be numerically calculated and plotted if so desired by the OP.

This can be physically demonstrated using a flexible chain suspended at D and having (A,B,E) points marked at given arc distances.

EDIT1

Mathematica platform numerically computed Catenaries with varying arc lengths, two initial slopes at D = (- 0.85,-1.12).