I swear, I wrote this post before he started referring to Greenland as "Iceland". It was meant to be hyperbole, but comedy is impossible.

Could you explain your thinking why it is similar to a root solve? I suppose you were thinking a polynomial interpolation and then inverse Vandermonde using the indices from the x-axis to generate a polynomial to invert. However, polynomial interpolation isn't generally monotonic, so that's a bust.

Here is a link to a video demonstrating how it remains broken, even whilst the switch remains on (timestamp 17:26): www.youtube.com/watch?v=Cq55...

Preface: I have no formal electrical training.
A circuit breaker has a sprung arm held by a small notch (magenta). If this notch is pushed by the solenoid (green) or bimetallic strip (blue), the arm swings out and keeps the contact point (red) broken until the switch is reset. Safe, but inadvisable.

This gives f(x)=(-2∓√(4+3x^2)-3x^2)/(3(2±√(4+3x^2))). Looking at the graph in Wolfram, it seems the positive-root gives you correct curve for R=1/2.

I took a stab at R=1/2. Using the compound-angle formula for 2/x=cot(2β/2)+cot(β/2), I got the polynomial 3cot(β/2)^2-(4/x)cot(β/2)-1=0, so cot(β/2)=(2±√(4+3x^2))/(3x). Putting this back into the compound-angle formula, cot(β)=(4±2√(4+3x^2)-3x^2)/(3x(2±√(4+3x^2))).

From the method I discussed in my last comment, you can get the result for R=P/Q by solving for 2/x=cot(Qβ/Q)+cot(Pβ/Q) as polynomials in cot(β/Q) and then solving for x, and then combining again via the compound-anfle formula. I suspect you would end up with a polynomial of degree max(P,Q).

However, I don't recognise any family of curves that matches. R=1 is a line, and R=2 is a hyperbola, but, given that conic sections are all degree-2 curves, this seems to fall apart for higher R. Sorry. 😕

This carries on up for higher values of R and, from what I see, each requires a solution to an R-degree polynomial.
It should be possible to find at least one real root for any R as rays extended for all α and β must intersect on the plane (if α!=π-β).

Using the compound angle formula for cotangent, we get 2/x=cot(β)+cot(2β)=cot(β)+((cot(β)^2-1)/(2cot(β))), which can be rearranged into a quadratic and solved as cot(β)=(2+√(4+3x^2))/(3x), and thus f(x)=(-1+√(4+3x^2))/3.
For R=3, you have to solve the cubic cot(β)^3-3cot(β)-(2/x)(3cot(β)^2-1)=0.

There is an explicit equation for each value of R, because f(x)=1-(x/tan(R*β(x)))=(x/tan(β(x)))-1, where 2/x=cot(R*β(x))+cot(β(x)). The problem comes for solving this second equation for arbitrary R.
If R=1, then it simplifies to f(x)=0. It starts to complicate for R=2. 🧵

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Socorro!

Many Stanley Kubrick films leave me feeling underwhelmed and bitterly dissatisfied. I have gained enjoyment from watching behind-the-scenes documentaries and reading the stories that he adapted and consuming analysis on the films. None of that translates into enjoyment of the presented product.