I'm looking for a formula (isn't everyone?).

this/these formula/e evaluate/s a position on the surface of a sphere, such that...

... the entire sphere is covered.

... the density of the sphere is equal at every point.

... the input is (prefferably but not neccesarily) in the range [0,1).

I have no idea where to start, any answer or even simply a start in the right direction would be wildly apreciated.

TIA

James

What follows requires two (rather than one) uniformly distributed random variables?

Define Sphericla coordinates as follows: the sphere has radius $r$; the 'longitude' is denoted by $\theta$ (and takes the range $[0,2\pi ]$); the latitude is denoted by $\phi$ and is in $[-\pi /2,\pi /2]$.

For a given $r$ and over all $\theta$, if $\phi$ lies in the range $[\phi ,\phi +\delta \phi ]$ then we are describing a 'ring' of the sphere with area

$2\pi r\cos (\phi )r\delta \phi =2\pi r^2\cos (\phi )\delta \phi$

Therefore the area covered for $\phi$ in $[-\pi /2,\phi ]$ is (integrating over $\phi$):

$A=2\pi r^2[\sin (\phi )+1]$

As an aside, if we set $\phi =\pi /2$, we get $A=4\pi r^2$ which is the familiar formula for the area of a sphere.

Now if we want a mapping between $\phi$ and $x$ (in [0,1]) such that the area is proportional to $\delta x$, then

$4\varphi r^{2}x=2\pi r^2[\sin (\phi )+1]$

$\phi ={\sin }^{-1}(2x-1)$

Finally, we observe that the area of a small piece of the sphere (described by $r$, $[\theta ,\theta +\delta \theta ]$, $[\phi ,\phi +\delta \phi ]$ is proportional to $\delta \theta$, therefore if $y$ is in [0,1] our second mapping is:

$\theta =2\pi y$

If you really must have just a single uniformly distributed variable, $z$, in [0,1] as an 'input' then I suppose you could construct the $x$, $y$ as being the odd and even digits of $z$ after the decimal point, but if you were ever going to write some sort of a computer program, it would make no sense.

Andre

That's really cool.

You seemed to brush over the proof that this has an uniform density. I'm sure it does and if you don't reply i'll go away beleiving it to be true, maybe even try to myself. It would just be satisfying to see a proof.

Thanks again for your time, it's much appreciated.

James

We constructed $x$, $y$ such that

(i) $x$ maps onto $\phi$ and $y$ maps onto $\theta$. There is no dependence of $\theta$ on $x$ etc.
(ii) The area bounded by $x$ in $[0,k]$ is proportional to $k$.

Alternatively, having constructed $x$, $y$ we can work the other way and compute the area between $(x,y)$ and $(x+\delta x,y+\delta y)$. We know that the area between $(r,\theta ,\phi )$ and $(r,\theta +\delta \theta ,\phi +\delta \phi )$ is given by:

$\mathrm{dA}=r^2\cos (\phi )\delta \phi \delta \theta$

and that

$\phi ={\sin }^{-1}(2x-1)$

$\theta =2\pi y$

implies

$\cos (\phi )\delta \phi =2\delta x$

$\delta \theta =2\pi \delta y$

substituting:

$\mathrm{dA}=r^{2}2\delta x2\pi \delta y=4\pi r^2\delta x\delta y$

The lack of this dependence of this expression on $x$, $y$, $\theta$, $\phi$ is one and the same as saying the area described on the sphere is proportional to the area described in the unit square.

As an aside, you could use this mapping to map the earth on a unit square. Such an area-preserving transformation is known as (guess what?) an ëqual area projection". If you search in Google for "Wolfram equal area projection" you'll see other ways of doing it. I can't see the mapping we described above.

Andre

excellent.

thanks again.