“袋子里的球体”平面到球面投影
我正在寻找数学变换，将 2D 平面 [0,1]x[0,1] 上的点变换到单位球面上。
最常见的投影是通过将 u 和 v 解释为球面坐标的角度来进行纬度-经度映射（将 u 映射到 [ 0,2PI] 和 v 到 [-PI/2, PI/2]）
这会在球体的极点上产生强烈的扭曲。人们可以将这种变换想象为将球体包裹到一张糖果纸中，并在两端旋转纸。这会在两端造成扭曲。
我正在寻找的转变可以是将球体放入纸的中间并将球体的所有侧面放在球体周围并将它们旋转到一个点上 - 这样你就得到了一个小纸袋，里面有你的球体。这会在“袋子”底部产生最小的变形，在顶部产生最大的变形 - 如果从下面看，变形在所有方向上都是相等的。
有人可以告诉我如何计算这种映射吗？
I'm looking for the mathematical transformation to transform points on a 2D plane [0,1]x[0,1] onto the unitsphere.
The most common projection would latitude-longitude mapping by interpreting u and v as the angles for the spherical coordinates (map u to [0,2PI] and v to [-PI/2, PI/2])
This gives strong distortions on the poles of the sphere. One can think of this transform as like wrapping the sphere into a bonbon-paper twirling the paper at the both ends. This will give distortions at those two ends.
The transformation I'm looking for can be rater thoght of putting the sphere into the middle of a paper and putting all sides around the sphere and twirl them together on a single spot - so you get a little paper-bag with your sphere in it. This yelds minimal distortion on the bottom of the "bag" and maximum distoriton on the top - and if seen from below, the distortion is equal in all directions.
Can someone tell me how to calculate this kind of mapping?
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对于您描述的映射，可以使用极坐标：(x,y)-->(r,alpha)，其中r在[0,1]中，表示距矩形O中心的距离之间的比率(0.5,0.5) 到当前点 P(x,y)，以及该线段在当前 alpha 值下可以具有的最大长度。然后将 r 映射到 [-PI/2, PI/2]，将 alpha 映射到 [0,2PI]。
For the mapping you describe, you can use polar coordinates: (x,y)-->(r,alpha), where r is in [0,1], representing the ratio between the distance from the center of the rectangle O(0.5,0.5) to the current point P(x,y), and the maximum length this segment could have at the current value of alpha. Then map r to [-PI/2, PI/2] and alpha to [0,2PI].
我认为您正在寻找指数地图。另请参阅使用离散指数图进行交互式贴花合成。
I think you are looking for the Exponential Map. See also Interactive Decal Compositing with Discrete Exponential Maps.
正确的答案取决于需要保留原始图像的哪些属性，因为每个不同的地图投影都会以不同的方式扭曲。有些保留区域，有些保留角度，有些保留距离。
假设案例与形状有关，那么我建议使用 Dymaxion 贴图，但请注意其平面表示不是完全的矩形。
有关其他选项，请参阅科罗拉多大学上的列表。
The right answer depends on which property of the original needs preserving because every distinct map projection distorts in a distinct way. Some preserve areas, some preserve angles, some preserve distances.
Assuming the case is about shapes, then I'd suggest a Dymaxion map but note that its planar representation is not fully rectangular.
For other options see the list at Colorado University.
如果您使用从 0 到 1 的 xy 轴（即第一象限）绘制问题草图，则使用相同的原点绘制轴从 0 到 pi/2 的投影第一个八分圆。从原点开始标记点(1,1)，则该点距离原点的大小为root(2)。现在您可以看到点 (1,1) 无法映射到球体上，因为它会出现在球体外部。
if you make a sketch of the problem using the x-y axes from 0 to 1 (ie first quadrant), then with the same origin draw the projected first octant with its axes from 0 to pi/2. Mark in point (1,1) from the origin then the magnitude of this point from the origin is root(2). You can now see that your point (1,1) cannot be mapped onto the sphere as it would appear outside it.
您可以检查 方位等距投影，但是，它将整个 2d 平面映射到球体上，并且不仅仅是一个矩形。也许有一种方法可以解决您的问题。
you can check the azimuthal equidistant projection, however, It maps the whole 2d plane on a sphere and not just a rectangle. Maybe there is a way to adjust for your problem.