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If a space has a differential volume element defined by:

$d\Omega=\sin^2(\alpha)\sin(\theta)d\alpha d\theta d\phi$

And $\alpha \in [0,\pi/2]$, $\theta \in [0,\pi]$, and $\phi \in [0,2\pi]$

How can I make a regular grid of points over the space that is uniform with respect to the the differential volume element?

I imagine a solution where I reparameterize the space in terms of, e.g., $u,v,w$, which are sampled uniformly. Then to get $\alpha,\theta,\phi$ I apply some transformation:

$\alpha = f(u)$

$\theta = g(v)$

$\phi = h(w)$

or something like that. How would I do this (i.e. find f, g, and h)?

share|improve this question
    
I doubt that it is possible, unless you really stretch the meaning of "regular grid". How would you put a "regular grid" on the 2-sphere? –  Robert Israel Sep 13 '11 at 18:54
    
@Robert Israel: you can do it by, e.g., $u \in [0,1]$ and $v \in [0,1]$, and then $\theta = 2\pi u$ and $\phi = \cos^{-1}(2v-1)$. The motivation behind this is for interpolation purposes. I want the error associated with linear interpolation to be the same everywhere, if I don't account for the curvature of the space then the error will be greater in certain regions, e.g. in the case of the 2-sphere the error will be greatest at the equator if one simply creates a grid where $\theta$ and $\phi$ are sampled uniformly over their domain. See the article "Sphere Point Picking" from mathworld. –  okj Sep 13 '11 at 19:16

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