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Given a sphere of radius $r$, and origin $x,y,z$ what is the simplest way I can generate an evenly distributed array of points on the sphere $(x_1,y_1,z_1),(x_2,y_2,z_2),\cdots(x_n,y_n,z_n)$.

Note I will be writing this as a function in Javascript, if it is any help.

EDIT

Essentially, I want to create a perfectly symmetrical shape with $X$ number of vertices that fits perfectly inside a sphere with radius $R$.

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marked as duplicate by Rahul, Macavity, user91500, hardmath, Najib Idrissi Jun 6 at 7:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Do you want a regular distribution of points, or do you want to pick the points randomly from a uniform distribution? –  Brian M. Scott Jan 15 '13 at 20:54
    
You can't equidistribute points on a sphere unless they are the corners of a Platonic solid, so $n=4,6,8,12,20$. Do you mean randomly distributed points with constant density? If the second, you could see mathworld or Google "random distribution on sphere" –  Ross Millikan Jan 15 '13 at 20:55
    
Random will work perfectly., really just whichever way gets the job done with the least amount of overhead to perform the calculation. Thanks @BrianM.Scott –  Jonathan Coe Jan 15 '13 at 20:56
1  
It won't be symmetric if the points are random. But it will "on average be symmetric" whatever that means. –  Ross Millikan Jan 15 '13 at 21:01
2  
If the aim is to pick random points from a uniform distribution on the sphere, shouldn't this be closed as a duplicate of How to find a random axis or unit vector in 3D? If the aim is to find a symmetrical set of equally spaced points on the sphere, see the answers under Which tessellation of the sphere yields a constant density of vertices? –  Rahul Jan 15 '13 at 21:20

3 Answers 3

up vote 2 down vote accepted

Use a uniform random number generator to generate an angle $\theta\in[0,2\pi)$ (essentially a longitude) and a $z\in[-1,1]$; the surface area cut by the planes $z=a$ and $z=b$ depends only on $|a-b|$, provided that $a,b\in[-1,1]$, so you get a uniform distribution.

Once you have $\theta$ and $z$, the point is $\left\langle\sqrt{1-z^2}\cos\theta,\sqrt{1-z^2}\sin\theta,z\right\rangle$ in rectangular coordinates.

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ We have to generate random points in $\ds{\pars{\theta,\phi}}$ with $\ds{0 \leq \theta < \pi}$ and $\ds{0 \leq \phi < 2\pi}$ in a spherical surface of radius $\ds{r}$ such that: $$ \vec{r} = x\,\hat{x} + y\,\hat{y} + z\,\hat{z}\,,\qquad x = r\sin\pars{\theta}\cos\pars{\phi}\,,\quad y = r\sin\pars{\theta}\sin\pars{\phi}\,,\quad z = r\cos\pars{\theta} $$ It means we choose evenly distributed solid angles $\ds{\Omega}$ at the sphere surface. Let's assume we have a generator of evenly distributed points $\ds{\braces{\xi}}$ in $\ds{\left[0, 1\right)}$. The probability distributions for $\ds{\theta}$ and $\ds{\phi}$ are given by $\ds{\half\,\sin\pars{\theta}}$ and $\ds{1 \over 2\pi}$, respectively. With a couple of values of $\ds{\xi}$ ( let's say $\ds{\xi_{\theta}}$ and $\ds{\xi_{\phi}}$ ) we'll have: $$ \int_{0}^{\theta}\half\sin\pars{t}\,\dd t=\xi_{\theta}\,,\qquad \int_{0}^{\phi}{1 \over 2\pi}\,\dd t=\xi_{\phi}\qquad\imp\qquad \left\lbrace\begin{array}{rcl} \theta & = &2\arcsin\pars{\root{\xi_{\theta}}} \\ \phi & = & 2\pi\xi_{\phi} \end{array}\right. $$

$$\large \color{#c00000}{\mbox{Below, we show a}\ \color{#000}{{\tt javascript}}\ \mbox{code that makes the job:}} $$ \begin{align} \end{align}

var TWOPI=2.0*Math.PI;

function randSphereSurface(r)// r: radius
{
 var phi=TWOPI*Math.random();
 var theta=2.0*Math.asin(Math.sqrt(Math.random()));
 var x=r*Math.sin(theta);
 var y=x*Math.sin(phi);
 x*= x*Math.cos(phi);
 var z=r*Math.cos(theta);

 return {"x":x,"y":y,"z":z};
}

Use as

var p=randSphereSurface(5.0); // sphere of radius 5.
document.write("The point is (", + p.x + "," + p.y + "," + p.z + ")");
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At the end of the Mathworld article it says you can generate three Gaussian random variables $x,y,z$. Then $r=\sqrt {x^2+y^2+z^2}$ and $\frac xr, \frac yr, \frac zr$ are equally distributed on the unit sphere.

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