Closest grid square to a point in spherical coordinates

Question

I am programming an algorithm where I have broken up the surface of a sphere into grid points (for simplicity I have the grid lines parallel and perpendicular to the meridians). Given a point A, I would like to be able to efficiently take any grid "square" and determine the point B in the square with the least spherical coordinate distance AB. In the degenerate case the "squares" are actually "triangles".

I am actually only using it to bound which squares I am searching, so I can also accept a lower bound if it is only a tiny bit off. For this reason, I need the algorithm to be extremely quick otherwise it would be better to just take the loss of accuracy and search a few more squares.

• I decided to repost this question to Math Overflow: https://mathoverflow.net/questions/854/closest-grid-square-to-a-point-in-spherical-coordinates. More progress has been made here

Answer 1

See Math Overflow: https://mathoverflow.net/questions/854/closest-grid-square-to-a-point-in-spherical-coordinates for an exact solution

Answer 2

For points on a sphere, the points closest in the full 3D space will also be closest when measured along the surface of the sphere. The actual distances will be different, but if you're just after the closest point it's probably easiest to minimize the 3D distance rather than worry about great circle arcs, etc.

To find the actual great-circle distance between two (latitidude, longitude) points on the sphere, you can use the first formula in this link.

Answer 3

A few points, for clarity.

Unless you specifically wish these squares to be square (and hence to not fit exactly in this parallel and perpendicular layout with regards to the meridians), these are not exactly squares. This is particularly visible if the dimensions of the square are big.

The question speaks of a [perfect] sphere. Matters would be somewhat different if we were considering the Earth (or other planets) with its flattened poles.

Following is a "algorithm" that would fit the bill, I doubt it is optimal, but could offer a good basis. EDIT: see Tom10's suggestion to work with the plain 3D distance between the points rather than the corresponding great cirle distance (i.e. that of the cord rather than the arc), as this will greatly reduce the complexity of the formulas.

Problem layout:  (A, B and Sq as defined in the OP's question)
 A  : a given point the the surface of the sphere
 Sq : a given "square" from the grid 
 B  : solution to problem : point located within Sq which has the shortest 
      distance to A.
 C  : point at the center of Sq

Tentative algorithm:
Using the formulas associated with [Great Circle][1], we can:
 - find the equation of the  circle that includes A and C
 - find the distance between A and C. See the [formula here][2] (kindly lifted
    from Tom10's reply).
 - find the intersect of the Great Circle arc between these points, with the
   arcs  of parallel or meridian defining the Sq.
   There should be only one such point, unless this finds a "corner" of Sq, 
   or -a rarer case- if the two points are on the same diameter (see 
   'antipodes' below).
Then comes the more algorithmic part of this procedure (so far formulas...):
 - find, by dichotomy, the point on Sq's arc/seqment which is the closest from
   point A.  We're at B! QED.

Optimization:  
 It is probably possible make a good "guess" as to the location
 of B, based on the relative position of A and C, hence cutting the number of
 iterations for the binary search.
 Also, if the distance A and C is past a certain threshold the intersection
 of the cicles' arcs is probably a good enough estimate of B.  Only when A
 and C are relatively close will B be found a bit further on the median or
 parallel arc in these cases, projection errors between A and C (or B) are
 smaller and it may be ok to work with orthogonal coordinates and their 
 simpler formulas.

Another approach is to calculate the distance between A and each of the 4 
corners of the square and to work the dichotomic search from two of these
points (not quite sure which; could be on the meridian or parallel...)

( * ) *Antipodes case*:  When points A and C happen to be diametrically 
opposite to one another, all great circle lines between A and C have the same
 length, that of 1/2 the circonference of the sphere, which is the maximum any
 two points on the surface of a sphere may be.  In this case, the point B will
 be the "square"'s corner that is the furthest from C. 

I hope this helps...

Answer 4

The lazy lower bound method is to find the distance to the center of the square, then subtract the half diagonal distance and bound using the triangle inequality. Given these aren't real squares, there will actually be two diagonal distances - we will use the greater. I suppose that it will be reasonably accurate as well.