Catenary equation in 3D

Question

I have two points with coordinates A(x1,y1,z1) and B(x2,y2,z2). There is a third point which is lowest point of the catenary curve. I only know z-coordinate of this third point. I need to find coordinates of the points that lie on this catenary that passes through these 3 points. I am interested in coordinates that lie in between A and B.

Answer 1

Shift (translate) both points together so that the first point lies at origin. Create all the needed coordinates between origin and displaced point. Rotate in the plane through $ \tan ^{-1} (y/x) $ so new anchor points are:

$$ (0,0,0) -(x_2-x_1, y_2-y_1,z_2-z_1)$$

Hope the catenary length is given and you know how to get the coordinates when endpoints at different $z$ coordinates are given. This done, shift the entire catenary back, bringing end points where they were held/fixed.

Answer 2

First apply a rotation around a vertical axis to let $y_1=y_2$, and subtract the given $z_a$ ($a$ for apex).

In these transformed coordinates, the equation of the catenary is

$$z=A\left(\cosh\left(\frac{x-x_a}A\right)-1\right)$$

with two unknown parameters $A$ and $x_a$.

With the two given points, we form the system

$$z_1=A\left(\cosh\left(\frac{x_1-x_a}A\right)-1\right),\\ z_2=A\left(\cosh\left(\frac{x_2-x_a}A\right)-1\right).$$

You can eliminate $x_a$ by

$$x_a=x_1-A\,\text{arcosh}\left(\frac{z_1}A+1\right)$$ and $$z_2=A\left(\cosh\left(\frac{x_2-x_1\,}A+\text{arcosh}\left(\frac{z_1}A+1\right)\right)-1\right).$$

This "appetizing" equation needs to be solved for $A$ by numerical methods.