The obvious solution of Lattitude & Longitude doesn't work because it generates points more densely near the poles, and the other thing I came up with (Pick a random point in the unit cube, if it's in the unit sphere map it to the surface, and restart if it's outside) doesn't always find a point within a fixed number of tries.

  • $\begingroup$ So what you want is a uniform distribution. It would be helpful to state this explicitly. $\endgroup$ – robjohn Apr 12 '19 at 0:30
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    $\begingroup$ Distribute longitude uniformly and the sine of the latitude uniformly. Then the distribution of points on the sphere will be uniform. $\endgroup$ – robjohn Apr 12 '19 at 0:32
  • $\begingroup$ @robjohn thank you, you're right that I forgot to specify that. $\endgroup$ – The Zach Man Apr 12 '19 at 2:50
  • $\begingroup$ "A uniformly distributed point" is an oxymoron, is it not? If not, then I suggest (r,theta,phi) = (1,0,0), which was drawn from my secret uniform distribution :P $\endgroup$ – Marius Ladegård Meyer Apr 12 '19 at 7:54
  • $\begingroup$ @MariusLadegårdMeyer: okay, I've adjusted the title further. I had tried to make as small a change that would make the uniform distribution clear. Of course, now the idea of a fixed number of random numbers per point is muddied. $\endgroup$ – robjohn Apr 12 '19 at 14:43

The Lambert cylindrical equal area projection maps the sphere to a cylinder, area to equal area. It is easy to generate a uniform distribution on a cylinder. Simply map it back to the sphere.

For $(u_1,u_2)$ uniform on $[0,1]^2$, either

$\mathrm{lat}=\arcsin(2u_1-1),\mathrm{lon}=2\pi u_2$


$z=2u_1-1,x=\sqrt{1-z^2}\cos(2\pi u_2),y=\sqrt{1-z^2}\sin(2\pi u_2)$


Your method, even though it doesn't finish in a fixed number of times, is a reasonable way to do it. Each trial succeeds with probability $\frac\pi6$, which is better than $\frac12$: the average number of trials is less than $2$.

Another standard method is to use the normal distribution. Generate $x, y, z$ independently from a standard normal distribution, then take the point $(x,y,z)$ and divide it by $\sqrt{x^2+y^2+z^2}$ as you did for points inside the cube. The multivariate normal distribution is rotationally symmetric, so this will get you evenly distributed points on the sphere.

(The Box–Muller transform is one way to generate normally distributed random numbers, and some versions of it do not use rejection sampling, so they can be done with a "fixed amount" of randomness.)


Is there a way to generate uniformly distributed points on a sphere from a fixed amount of random real numbers per point?

This Stack Overflow answer mentions the Fibonacci sphere algorithm, and the currently unanswered Math SE question Is the Fibonacci lattice the very best way to evenly distribute N points on a sphere? So far it seems that it is the best? contains several excellent links in the question and in comments.

The method is deterministic, using either zero or one random number, no matter the number of points generated.

Florida State University's John Burkardt's web page Fibonacci Spiral Grid on a Sphere provides some references and examples of code implementations.

Edward Saff, Arno Kuijlaars, Distributing Many Points on a Sphere, The Mathematical Intelligencer, Volume 19, Number 1, 1997, pages 5-11.

Richard Swinbank, James Purser, Fibonacci grids: A novel approach to global modelling, Quarterly Journal of the Royal Meteorological Society, Volume 132, Number 619, July 2006 Part B, pages 1769-1793.

Extreme Learning's Evenly distributing points on a sphere addresses several criteria for judging the level of uniformity provided by different solutions, and describes the process of first generating a Fibonacci grid and then transforming it to a sphere.

An example of these Fibonacci Grids is shown below. These points sets can be transformed to a well-known Fibonacci spirals via the transformation

enter image description here

I can't be sure it will be sufficiently uniform for you, but it certainly looks nice!

enter image description here Source

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    $\begingroup$ That's not really what the question was asking for, but is useful nevertheless, thank you. $\endgroup$ – The Zach Man Apr 18 '19 at 2:29
  • $\begingroup$ @TheZachMan in regards to "what the question was asking for" I've certainly address the title of your question which contains the only instance of a question mark in your post, and there's nothing in the body of your question that restricts answers only to "the obvious solution of Lattitude & Longitude". Your other question was much more carefully constrained. $\endgroup$ – uhoh Apr 18 '19 at 2:34
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    $\begingroup$ Ah, I see what you mean - the original title was a bit unclear, but it seems the edited version is unclear in a different way, I'll add a word or two to clarify. $\endgroup$ – The Zach Man Apr 18 '19 at 2:42

You can still use the longitude/latitude method. You just need to transform the latitude such that the probability of falling in any given latitude band is proportional to its surface area.

The length of the circumference at a latitude, l, is proportional to cos(l). (0 at -90 degrees, max at the equator, 0 at +90 degrees). Therefore, by integrating we know that the total surface area between any two latitudes, l1 and l2, is proportional to sin(l2) - sin(l1).

By inverting that, we arrive at a formula for transforming a [0, 1] uniform random variable (X) into a latitude such that any point on the sphere is equally likely to be selected:

latitude = arcsin(2X - 1)

Your second method can be improved in this way: Select the three coordinates (x, y, z) in a linear distribution from -1 to 1.

If (y * y + z * z) > 1, select replacements for these two only. (You will only have to re-roll them 4-pi/4 (roughly 21.4%) of the time; average random numbers used is 1 + 8/pi = ~ 3.546.

Keep x as it is, scale both y and z by sqrt((1 - x * x)/(y * y + z * z)).


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