Thread by @TamasGorbe

What's the shape of a hanging chain? Galileo knew that it's something like a parabola, but not quite. This curve is called the catenary and its the graph of a hyperbolic cosine function cosh(x)=(eˣ+e⁻ˣ)/2. [Wiki bit.ly/2RmyP15] #50FamousCurves

Here's a derivation via Noether's theorem. tamasgorbe.wordpress.com/2015/05/27/wha…

Fun fact: Smooth travel on square wheels is possible as long as the road consists of upside down catenary arcs (the length of which equals the side length of the wheel). [Source: ] #50FamousCurves

Fun fact: The word "catenary" comes from the Latin catēna meaning "chain". #50FamousCurves

Fun fact: Simple suspension bridges - often seen in adventure movies - are hyperbolic cosines (catenaries). [Source: AF Archive Alamy] #50FamousCurves

How to draw a catenary?
1) Draw a tractrix (as previously seen on #50FamousCurves).
2) Draw as many normals to the tractrix as you can.
The envelope of these lines is a catenary. #50FamousCurves

Fun fact: The Gateway Arch in St. Louis, Missouri is an upside down weighted catenary whose equation is y=a-b cosh(x/c), where a≈211.5 m, b≈21 m and c≈30.4 m. #50FamousCurves

Fun fact: Soap film stretched between two circular wires takes the form of a catenoid. This is the surface you get, if you rotate a catenary about its directrix. It's one of the two minimal surfaces of revolution (the plane being the other one). #50FamousCurves

The area of a surface of revolution can be calculated via the integral formula
A=2π ∫ y(x)√[1+y'(x)²] dx.

For a catenoid stretched between wires of radii R & at distance D apart, we have -D/2<x<D/2 and y(x)=a ch(x/a) such that ch(D/2a)=R/a. This yields
A=πa²sh(D/a)+πaD.
(1/5)

If the wires are too far apart, the soap film collapses into two separate disks. The critical distance D* is reached, when the surface area A equals the total area of the two disks, 2R²π.

This yields the equation
πa²sh(D*/a)+πaD*=2R²π.

Let's solve this for D*!
(2/5)

Aim: Solve πa²sh(D*/a)+πaD*=2R²π for D*.

Dividing both sides by 2πa² gives
sh(D*/a)/2+D*/2a=(R/a)².
Using ch(D*/2a)=R/a & sh(2x)=2sh(x)ch(x) yields
sh(D*/2a)ch(D*/2a)+D*/2a=ch²(D*/2a).
Let x=D*/2a & ÷ by ch(x) to get
sh(x)+x/ch(x)=ch(x).
It's a transcendental equation.
(3/5)

Subtract sh(x) from both sides of
sh(x)+x/ch(x)=ch(x)
to get
x/ch(x)=ch(x)-sh(x).
Use the definitions ch(x)=(eˣ+e⁻ˣ)/2 and sh(x)=(eˣ-e⁻ˣ)/2 to obtain
x/ch(x)=e⁻ˣ.
Multiply both sides by eˣch(x) to get
xeˣ=ch(x).
Use WolframAlpha to find the approx. solution
x≈0.639232.
(4/5)

For the critical distance, we have
D*/2a≈0.639232,
hence
R/a=ch(D*/2a)≈1.211361.
The fraction of these two lines is
D*/2R≈0.527697.

Therefore the critical distance of wires is
D*≈1.055395 R,
where R is the radii of the loops.
(5/5)
Inspired by @RobJLow. Thanks!