分割单位球体 [英] Partition the Unit Sphere

问题描述

我想为单位向量制作查询表。每个单位矢量将映射到该表中的一个bin，并且该bin将包含一些具有相似方向的一组向量的信息。

我可以很容易地表示矢量使用（$ \theta $，$ \phi $，1），然后将角度范围切分成二维查找表（所以第一个bin是在0到$ 2范围内的theta * \pi / N $，其中N是我为theta方向所需的箱数）。这样做的麻烦在于，我认为一些箱子将在单位球体的表面上代表更大的区域，而我希望得到大小基本相同的区域。

我错误地认为均匀划分角度范围会使一些箱子比其他箱子大一些？如果没有，是否有人知道制作这个查找表的更好方法？

我发现了一些文章和演讲，比如这一个，但我不会说谎，我不明白它（我听说过勒贝格的措施，但如果我知道它是什么，我会被诅咒的）如果你将经度分成，那么它就不会有什么特别的前景。

然后，为了在单位球体上获得相同的面积域，沿纬度维度必须有不均匀的分段。
两条恒定纬度线（平行线）之间的球形区域仅取决于在高度上，即该部分到垂直轴线的投影长度。
这意味着如果你将垂直轴分成相等长度的部分，那么你将把球体分割成相同的区域。

底线是：以下 N * M 域有相同的区域：

• 2 * k * pi / N <结果， 2 *（k + 1）pi / N ， k = 0 ... N-1


• -1 + 2 * j / M < sin（纬度） -1 + 2 *（j + 1）/ M ， j = 0 ... M-1

I'd like to make a lookup table for unit vectors. Each unit vector would map to a bin in this table, and the bin would contain some information for a small set of vectors with similar directions.

I could easily represent a vector using ($\theta$, $\phi$, 1), and then chop the angle ranges into bins to make a 2D lookup table (so the first bin is theta in the range of 0 to $2*\pi / N$, where N is the number of bins I want for the theta direction). The trouble with this is that I think that some bins are going to represent larger areas on the surface of the unit sphere than others, and I would like to get regions of roughly the same size.

Am I wrong in thinking that evenly dividing the angle range would make some bins larger than others? If not, does anyone know a better way of making this lookup table?

I found some papers and presentations like this one, but I'm not going to lie, I don't understand it (I've heard of Lebesgue measure, but I'll be damned if I know what it means), and it doesn't look particularly promising anyway.

解决方案

If you split the longitude into N equal size segments, then, to get equal area domains on the unit sphere, you will have to have "uneven" segments along the latitude dimension. The area of the spherical segment between two lines of constant latitude (parallels) only depends on the "height", i.e., the length of the projection of the segment to the vertical axis. This means that if you split the vertical axis into equal length parts, then you will be splitting the sphere in to equal area domains.

The bottom line is: the following N*M domains have equal areas:

• 2*k*pi/N < longitute < 2*(k+1)pi/N, k=0...N-1
• -1 + 2*j/M < sin(latitude) < -1 + 2*(j+1)/M, j=0...M-1

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