Chamberlin trimetric projection

[2] As originally implemented, the projection algorithm begins with the selection of three base points to form a spherical triangle minimally enclosing the area to be mapped.

Chamberlin did not specify how to handle this case, but it would be determined by which definition of triangle center is chosen, as noted next.

In the remaining case, which is most of the map, connecting the three points of intersection of the circles by line segments creates a small triangle.

Rather, the projection was conceived to minimize distortion of distances everywhere with the side-effect of balancing between areal equivalence and conformality.

[3] This projection is not appropriate for mapping the entire sphere because the outer boundary would loop and overlap itself in most configurations.