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I am in the following situation:

I have a bunch of points $(r_i)$ on the sphere given by their angles $(\theta,\varphi).$ The radius $R$ of the sphere is also known.

The points are all in one part of the sphere. Coordinates of sights in NYC might be a good example.

So now rather than talking the positions in $(\theta,\varphi)$, I would like to find a local cartesian coordinate system such that I can express my coordinates as $(x_i,y_i)$.

In particular, if I take the euclidean distance between my points, then this should correspond to the actual distance.

I understand how to go from $(\theta,\varphi)$ to $(x,y,z)$ on the entire sphere, but here I would like to have a local chart and omit one of the three coordinates.

J.Doe
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  • Removing a coordinate sounds like a projection to me. The question would be how to choose a plane to project onto, such that the errors from curved sphere to flat plane are minimal. – mvw Jul 04 '21 at 10:40
  • As radius is fixed, then you already have plane $(\theta,\varphi)$ with 2 cartesian coordinates. Why you need some other? – zkutch Jul 04 '21 at 11:13
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    @zkutch If I have two points $(\theta_1,\varphi_1)$ and $(\theta_2,\varphi_2)$ then $\sqrt{(\theta_1-\theta_2)^2+ (\varphi_1-\varphi_2)^2}$ is not the actual distance between the points, even if they are close on the sphere. Therefore, this coordinate system does not satisfy this requirement. – J.Doe Jul 04 '21 at 11:28
  • How big is the local region, and how accurate do you want the distance to be? You might find this recent related question helpful: https://math.stackexchange.com/q/4178653/207316 – PM 2Ring Jul 04 '21 at 11:32
  • If you need more accuracy than a spherical Earth model gives, see https://en.wikipedia.org/wiki/Geographical_distance – PM 2Ring Jul 04 '21 at 11:42
  • @PM2Ring the leading order taylor expansion should be the correct distance and the area I am considering is something like the city of New York. One might argue that for distances there, the curvature of the earth is negligible – J.Doe Jul 04 '21 at 11:50
  • Sure, you can use a plane approximation on that distance scale. You just need to scale the longitudinal distance appropriately. It's simple to do that if we assume the Earth is a sphere, as in the question I previously linked. You can divide the longitudinal distance by $\cos(\varphi)$, where $\varphi$ is the usual (geodetic) latitude. But you get better accuracy if you use an ellipsoidal model, because the length of a parallel of latitude isn't exactly $2\pi R \cos(\varphi)$, where $R$ is the mean radius of the Earth. – PM 2Ring Jul 04 '21 at 12:06
  • IIRC, there does not exist a coordinate system on even a small region of a sphere in which spherical distance takes exactly the form of Cartesian distance. That's an immediate consequence of Gauss's Theorema Egregium. There does, however, exist an explicit (but slightly messy) trigonometric formula for spherical distance in terms of latitude and longitude. – Andrew D. Hwang Jul 04 '21 at 15:06
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    See also this related answer: https://math.stackexchange.com/a/2611287/255730 – Intelligenti pauca Jul 04 '21 at 16:24
  • There are many ways to solve your problem: orthographic projection onto a plane tangent to the Earth, stereographic projection, central projections onto various cylinders, central projection onto a cone, and projections by mathematical formula without such physical analogues, such as Mercator projection or the projection mentioned in the previous comment. All give good approximations of distance over a small area, often while preserving some other property exactly. – David K Dec 12 '21 at 16:29

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