Polar coordinates on sphere

Let xc\boldsymbol{x}_c be a point on S2 \mathbb{S}^2. This induces a notion of greatcircle distance, which is the length of a geodesic connecting some arbitrary point x∈S2 \boldsymbol{x} \in \mathbb{S}^2. For calculations we mostly use cartesian coordinates to reduce the algebra we need to do. This distance can be calculated as d(xc,x)=arccos⁡(xc⋅x) d(\boldsymbol{x}_c, \boldsymbol{x}) = \arccos(\boldsymbol{x}_c \cdot \boldsymbol{x}).

For the purposes of constructing high-quality topography that does not contain lat-lon artifacts (like in our BWTopo paper), I want to introduce the notion of phase, assuming we have some fixed xc \boldsymbol{x}_c that is the center of our coordinate system and some xo≠−xc \boldsymbol{x}_o \neq -\boldsymbol{x}_c which induces a reference geodesic with respect to which we can calculate the phase of any point on Y≡S2∖{xc,−xc}\mathbb{Y} \equiv \mathbb{S}^2 \setminus \{\boldsymbol{x}_c, -\boldsymbol{x}_c\}.

Let x∈Y\mathbb{x} \in \mathbb{Y}. Then the triple (xc,xo,x)(\boldsymbol{x}_c, \boldsymbol{x}_o, \boldsymbol{x}) forms a triangle on the sphere, which we can use to do trigonometry. Our phase is essentially∢xoxcx\sphericalangle \boldsymbol{x}_o \boldsymbol{x}_c \boldsymbol{x} . Let xo⊥\boldsymbol{x}_o^\perp be the point resulting from rotating xc\boldsymbol{x}_c 90º ccw about xo\boldsymbol{x}_o. This can be constructed as R=[ux2(1−cos⁡θ)+cos⁡θuxuy(1−cos⁡θ)−uzsin⁡θuxuz(1−cos⁡θ)+uysin⁡θuxuy(1−cos⁡θ)+uzsin⁡θuy2(1−cos⁡θ)+cos⁡θuyuz(1−cos⁡θ)−uxsin⁡θuxuz(1−cos⁡θ)−uysin⁡θuyuz(1−cos⁡θ)+uxsin⁡θuz2(1−cos⁡θ)+cos⁡θ] R={\begin{bmatrix}u_{x}^{2}\left(1-\cos \theta \right)+\cos \theta &u_{x}u_{y}\left(1-\cos \theta \right)-u_{z}\sin \theta &u_{x}u_{z}\left(1-\cos \theta \right)+u_{y}\sin \theta \\ u_{x}u_{y}\left(1-\cos \theta \right)+u_{z}\sin \theta &u_{y}^{2}\left(1-\cos \theta \right)+\cos \theta &u_{y}u_{z}\left(1-\cos \theta \right)-u_{x}\sin \theta \\ u_{x}u_{z}\left(1-\cos \theta \right)-u_{y}\sin \theta &u_{y}u_{z}\left(1-\cos \theta \right)+u_{x}\sin \theta &u_{z}^{2}\left(1-\cos \theta \right)+\cos \theta \end{bmatrix}}

The sign of xo⊥⋅x \boldsymbol{x}_o^\perp \cdot \boldsymbol{x} indicates whether to add π \pi to ∢xoxcx\sphericalangle \boldsymbol{x}_o \boldsymbol{x}_c \boldsymbol{x} Spherical trig should give the angle value.

What then happens if we let x=rcos⁡θ,y=rsin⁡θ x = r\cos\theta, y=r\sin\theta? This still has lat-lon issues, but we've made a coordinate mullet that hides these problems at the antipode. For, e.g., compact topography, this is acceptable. Todo: make a colab notebook exploring how these coordinates look in practice.

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