Given a point cloud on a sphere this Python package computes center and radius of the smallest (spherical) circle enclosing the point cloud. The algorithm takes linear time and automatically detects whether the point cloud is contained in a hemisphere. If so, the smalles enclosing circle is returned. Else, an exception is raised.
See A simple linear time algorithm for smallest enclosing circles on the (hemi)sphere for details on the algorithm.
Run
pip install secots
or copy secots.py from this repo to your working directory.
Put your point cloud into a list of longitude-latitude pairs or into a corresponding (n, 2)-shaped NumPy array. Then call secots.smallest_circle(points). This will return three floats: longitude and latitude of the circle's center as well as the radius. Example:
import secots
points = [
(-30, 10),
(0, 20),
(20, -40),
(10, 10),
(20, 30)
]
lon, lat, r = secots.smallest_circle(points)
print(f'center: ({lon}, {lat}), radius: {r}')More examples are in the repo's example directory.
The smallest_circle function raises NotHemisphereError if the point cloud is not contained in a hemisphere. In this case the algorithm is not able to compute a smallest enclosing circle. See above mentioned paper for details.
The smallest_circle function accepts an additional keyword argument hemitest. Use hemitest=False if you are very sure that your point cloud is contained in a hemisphere. This saves (an almost negligible amount of) computation time. Example:
lon, lat, r = secots.smallest_circle(points, hemitest=False)Open a GitiHub issue for bug reports. File pull requests against the dev branch.
The dev branch contains the code for next release, whereas the main branch holds the current release.