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Foreword

This question was inspired by initial mistakes in this question. I wanted to explore the strange circle with $A>\pi r^2$ and got lost into geometrical jungle.

A spherical cap is usually described by it's height $h$, radius of the base $a$ and radius of the sphere $r$. Besides we have a relation $a^2+(r-h)^2=r^2$ so the cap actually only has two degrees of freedom.

When considering a case of circle on a sphere (like the Earth), it's the surface of a spherical cap, but we can't see the height or other classic parameters of this spherical cap. Instead we encounter parameters like area $A$, perimeter $C$ and diameter $d$. I will try to write formulae for these: $$A=2\pi rh$$ $$C=2\pi a$$ $$d=2r\theta$$

Where $\theta$ is introduced with $\sin\theta=\frac{a}{r}$ or $\cos\theta=(1-\frac{h}{r})$.

As the cap only has two degrees of freedom, it should be a good problem with unique solution if only two values are known.

A-C Problem

The simple example: $A$ and $C$ are given. I show this only as example before the other tasks.

We can easily find $a=\frac{C}{2\pi}$ and then solve:

$$A=2\pi rh = 2\pi r (r-\sqrt{r^2-a^2})$$

to obtain

$$r=\frac{A}{2\sqrt{\pi(A-\pi a^2)}}$$

Parameters $a$ and $r$ are now known and it is straightforward to find $h$, $d$, $\theta$ and anything else.

C-d Problem

$C$ and $d$ are given.

First step once again is easy: $a=\frac{C}{2\pi}$. The knowing of $a$ suggests that we might prefer the $\theta$ definition with the sine. We obtain: $$d=2r\theta=2r\arcsin\frac{a}{r}$$

That's where I got stuck. Is there an easy way to solve this? Is there a hard way to solve this?

I tried solving numerically with Wolfram Mathematica and it went into complex numbers. When solving the geometrically equivalent $\sin\frac{d}{2r}=\frac{a}{r}$ I got some roots, but the search seemed very unstable and not finding the root closest to initial value. Besides it was finding a lot of different roots while I am pretty sure that there should only be one $r$ for each pair of $d$ and $a$. What would be the right formulation to solve this problem numerically?

A-d problem

$A$ and $d$ are given. This seems the hardest one. We obtain a system of equations:

$$A=2\pi r h$$ $$d=2r\arccos\left(1-\frac{h}{r}\right)$$

(The first equation suggested solving the problem in $h$ and $r$).

Is this system solvable analytically? Or numerically? My experiments with Wolfram Mathematica never converged.

Of course, I could express $h$ or $r$ from the first equation and leave only the second in one of these forms: $$d = 2r\arccos\left(1-\frac{A}{2\pi r^2}\right)$$ $$d = \frac{A}{\pi h}\arccos\left(1-\frac{2\pi h^2}{A}\right)$$

But I have no idea how to solve these either.

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1 Answer 1

up vote 1 down vote accepted

For the C-d problem, I would try letting $x=\frac1r$ and then solving $\sin(\frac{dx}2)=ax$ as this is likely to be much more numerically stable. And by using the first two terms of the Taylor series for sin you can get a good approximation:- $$\sin(\frac{dx}2)\approxeq \frac{dx}2-\frac{d^3x^3}{3!2^3}$$ using this approximation let us solve $$ax=\frac{dx}2-\frac{d^3x^3}{3!2^3}$$ $$a=\frac d2-\frac{d^3x^2}{48}$$ $$d^3x^2=24(d-2a)$$ $$x=\sqrt{\frac{24(d-2a)}{d^3}}$$ so $$r\approxeq\sqrt{\frac{d^3}{24(d-2a)}}$$

Likewise for the A-d problem we can turn $$d=2r\arccos(1-\frac A{2\pi r^2})$$ $$\cos(\frac d{2r})=1-\frac A{2\pi r^2}$$ $$\cos(\frac{dx}2)=1-\frac{Ax^2}{2\pi}$$ Then using the approximation $\cos(y)\approxeq 1-\frac{y^2}{2!}+\frac{y^4}{4!}$ $$1-\frac{d^2x^2}{2!2^2}+\frac{d^4x^4}{4!2^4}=1-\frac{Ax^2}{2\pi}$$ $$-\frac{d^2}{8}+\frac{d^4x^2}{384}=-\frac{A}{2\pi}$$ $$d^4x^2=384\left(\frac{d^2}8-\frac A{2\pi}\right)$$ $$x^2=\frac{3072\pi(d^2\pi-4A)}{d^4}$$ so $$r\approxeq\frac{d^2}{\sqrt{3072\pi(d^2\pi-4A)}}$$

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