# Do the endpoints of a catenary have to be horizontally aligned?

Nothing I see in the definition of a catenary says this must be the case, but every illustration I've seen draws it that way. I'm assuming that a cable hanging from two points at different heights off the ground is still a catenary but how does that change the equation and properties?

• Any hanging chain will form a catenary curve. It is the most stable / energy-efficient configuration. Commented Dec 4, 2023 at 7:06

Endpoints can be at different heights. One way to derive the equation is by minimizing the total potential energy of the cable, which gives an ODE via Euler-Lagrange equations. The general solution of that ODE is (a scaled version of) $$\cosh(x)$$, and only then are the boundary conditions imposed in order to fix the parameters of the general solution. This will not change the fact that the catenary is (a part of) an appropriately scaled hyperbolic cosine. See Wikipedia for the derivation.

In its most general form, the catenary will always be of the form $$a\cosh(\frac{x-b}a)+c$$ with appropriate constants $$a,b,c\in\mathbb R$$, and the endpoints will determine those constants.

• To determine 4 constants $(a,b,c,d) ,$ it is seen that the catenary passes through two end points. What are the other 2 conditions? Commented Dec 4, 2023 at 16:49
• Good question - indeed there are only three constants (I have edited my answer). The length of the cable would be the third condition needed: Given two endpoints in space, there are different catenary curves going through them, depending on their lengths. Commented Dec 4, 2023 at 17:24
• OK. Apart from the arc length many associated geometric constraints/conditions like slopes, rotation of chain between the 2 points, area under curve etc. can also be chosen as third condition right? Commented Dec 5, 2023 at 0:05
• Yes (not sure what you mean by rotation, though) - a slope at one of the endpoints, or the area enclosed by the chain would be other conditions that would allow to determine the 3rd parameter. Commented Dec 5, 2023 at 7:56
• Angle between tangents at A and B. Commented Dec 5, 2023 at 15:40

Not necessary, it will be clear when you handle the derivation leading to its ODE.

While integrating the ODE we can start with any slope at A as initial condition and end up at any slope at B.

Consider equilibrium of forces in a triangle. $$q$$ is weight per unit length of arc or cable represented by a downward arrow. Vertical force or weight upto any point is $$\int q~ds$$, a variable. Horizontal force at vertex V is a constant H.

$$u = \frac{dy}{dx}= \tan \phi= \dfrac{\int q ~ ds}{H}$$

Differentiate w.r.t. x

$$\frac{du}{dx}=\frac{q}{H} ~\sec \phi$$

$$\frac{du/dx}{\sqrt{1+u^2}}=\frac{q}{H}= \frac{1}{c} \tag1$$

Integrating

$$u= \sinh\frac{x}{c} +s_A$$

with an arbitrary initial slope $$s_A$$. Further integrate

$$y= c \cosh \frac{x}{c} +s_A \cdot x \tag2$$

with initial condition $$x=0,~y=c$$ at A, the unsymmetric catenary is sketched below.

Special case $$s_A=s_V =0$$ is commonly presented in a nice symmetric situation,

$$y = c \cosh \frac{x}{c}$$

Note that $$c$$ represents the radius curvature at lowest point vertex V.

Numerical

With input initial conditions $$(x_A,y_A)= (-1,2) ; (x_B,y_B)= (2,4);$$

Computed Output as graph for equation 2):

$$u= \tan (x/c), ~~ y= c \log(\sec (x/c))$$