Locating the centroid (center of mass) of spherical polygons

Anybody finding this, make sure to check Don Hatch's answer which is probably better.

I think this will do it. You should be able to reproduce this result by just copy-pasting the code below.

• You will need to have the latitude and longitude data in a file called longitude and latitude.txt. You can copy-paste the original sample data which is included below the code.
• If you have mplotlib it will additionally produce the plot below
• For non-obvious calculations, I included a link that explains what is going on
• In the graph below, the reference vector is very short (r = 1/10) so that the 3d-centroids are easier to see. You can easily remove the scaling to maximize accuracy.
• Note to op: I rewrote almost everything so I'm not sure exactly where the original code was not working. However, at least I think it was not taking into consideration the need to handle clockwise / counterclockwise triangle vertices.

Legend:

• (black line) reference vector
• (small red dots) spherical triangle 3d-centroids
• (large red / blue / green dot) 3d-centroid / projected to the surface / projected to the xy plane
• (blue / green lines) the spherical polygon and the projection onto the xy plane

from math import *try:    import matplotlib as mpl    import matplotlib.pyplot    from mpl_toolkits.mplot3d import Axes3D    import numpy    plotting_enabled = Trueexcept ImportError:    plotting_enabled = Falsedef main():    # get base polygon data based on unit sphere    r = 1.0    polygon = get_cartesian_polygon_data(r)    point_count = len(polygon)    reference = ok_reference_for_polygon(polygon)    # decompose the polygon into triangles and record each area and 3d centroid    areas, subcentroids = list(), list()    for ia, a in enumerate(polygon):        # build an a-b-c point set        ib = (ia + 1) % point_count        b, c = polygon[ib], reference        if points_are_equivalent(a, b, 0.001):            continue  # skip nearly identical points        # store the area and 3d centroid        areas.append(area_of_spherical_triangle(r, a, b, c))        tx, ty, tz = zip(a, b, c)        subcentroids.append((sum(tx)/3.0,                             sum(ty)/3.0,                             sum(tz)/3.0))    # combine all the centroids, weighted by their areas    total_area = sum(areas)    subxs, subys, subzs = zip(*subcentroids)    _3d_centroid = (sum(a*subx for a, subx in zip(areas, subxs))/total_area,                    sum(a*suby for a, suby in zip(areas, subys))/total_area,                    sum(a*subz for a, subz in zip(areas, subzs))/total_area)    # shift the final centroid to the surface    surface_centroid = scale_v(1.0 / mag(_3d_centroid), _3d_centroid)    plot(polygon, reference, _3d_centroid, surface_centroid, subcentroids)def get_cartesian_polygon_data(fixed_radius):    cartesians = list()    with open('longitude and latitude.txt') as f:        for line in f.readlines():            spherical_point = [float(v) for v in line.split()]            if len(spherical_point) == 2:                spherical_point.append(fixed_radius)            cartesians.append(degree_spherical_to_cartesian(spherical_point))    return cartesiansdef ok_reference_for_polygon(polygon):    point_count = len(polygon)    # fix the average of all vectors to minimize float skew    polyx, polyy, polyz = zip(*polygon)    # /10 is for visualization. Remove it to maximize accuracy    return (sum(polyx)/(point_count*10.0),            sum(polyy)/(point_count*10.0),            sum(polyz)/(point_count*10.0))def points_are_equivalent(a, b, vague_tolerance):    # vague tolerance is something like a percentage tolerance (1% = 0.01)    (ax, ay, az), (bx, by, bz) = a, b    return all(((ax-bx)/ax < vague_tolerance,                (ay-by)/ay < vague_tolerance,                (az-bz)/az < vague_tolerance))def degree_spherical_to_cartesian(point):    rad_lon, rad_lat, r = radians(point[0]), radians(point[1]), point[2]    x = r * cos(rad_lat) * cos(rad_lon)    y = r * cos(rad_lat) * sin(rad_lon)    z = r * sin(rad_lat)    return x, y, zdef area_of_spherical_triangle(r, a, b, c):    # points abc    # build an angle set: A(CAB), B(ABC), C(BCA)    # http://math.stackexchange.com/a/66731/25581    A, B, C = surface_points_to_surface_radians(a, b, c)    E = A + B + C - pi  # E is called the spherical excess    area = r**2 * E    # add or subtract area based on clockwise-ness of a-b-c    # http://stackoverflow.com/a/10032657/377366    if clockwise_or_counter(a, b, c) == 'counter':        area *= -1.0    return areadef surface_points_to_surface_radians(a, b, c):    """build an angle set: A(cab), B(abc), C(bca)"""    points = a, b, c    angles = list()    for i, mid in enumerate(points):        start, end = points[(i - 1) % 3], points[(i + 1) % 3]        x_startmid, x_endmid = xprod(start, mid), xprod(end, mid)        ratio = (dprod(x_startmid, x_endmid)                 / ((mag(x_startmid) * mag(x_endmid))))        angles.append(acos(ratio))    return anglesdef clockwise_or_counter(a, b, c):    ab = diff_cartesians(b, a)    bc = diff_cartesians(c, b)    x = xprod(ab, bc)    if x < 0:        return 'clockwise'    elif x > 0:        return 'counter'    else:        raise RuntimeError('The reference point is in the polygon.')def diff_cartesians(positive, negative):    return tuple(p - n for p, n in zip(positive, negative))def xprod(v1, v2):    x = v1[1] * v2[2] - v1[2] * v2[1]    y = v1[2] * v2[0] - v1[0] * v2[2]    z = v1[0] * v2[1] - v1[1] * v2[0]    return [x, y, z]def dprod(v1, v2):    dot = 0    for i in range(3):        dot += v1[i] * v2[i]    return dotdef mag(v1):    return sqrt(v1[0]**2 + v1[1]**2 + v1[2]**2)def scale_v(scalar, v):    return tuple(scalar * vi for vi in v)def plot(polygon, reference, _3d_centroid, surface_centroid, subcentroids):    fig = mpl.pyplot.figure()    ax = fig.add_subplot(111, projection='3d')    # plot the unit sphere    u = numpy.linspace(0, 2 * numpy.pi, 100)    v = numpy.linspace(-1 * numpy.pi / 2, numpy.pi / 2, 100)    x = numpy.outer(numpy.cos(u), numpy.sin(v))    y = numpy.outer(numpy.sin(u), numpy.sin(v))    z = numpy.outer(numpy.ones(numpy.size(u)), numpy.cos(v))    ax.plot_surface(x, y, z, rstride=4, cstride=4, color='w', linewidth=0,                    alpha=0.3)    # plot 3d and flattened polygon    x, y, z = zip(*polygon)    ax.plot(x, y, z, c='b')    ax.plot(x, y, zs=0, c='g')    # plot the 3d centroid    x, y, z = _3d_centroid    ax.scatter(x, y, z, c='r', s=20)    # plot the spherical surface centroid and flattened centroid    x, y, z = surface_centroid    ax.scatter(x, y, z, c='b', s=20)    ax.scatter(x, y, 0, c='g', s=20)    # plot the full set of triangular centroids    x, y, z = zip(*subcentroids)    ax.scatter(x, y, z, c='r', s=4)    # plot the reference vector used to findsub centroids    x, y, z = reference    ax.plot((0, x), (0, y), (0, z), c='k')    ax.scatter(x, y, z, c='k', marker='^')    # display    ax.set_xlim3d(-1, 1)    ax.set_ylim3d(-1, 1)    ax.set_zlim3d(0, 1)    mpl.pyplot.show()# run it in a function so the main code can appear at the topmain()

Here is the longitude and latitude data you can paste into longitude and latitude.txt

  -39.366295      -1.633460  -47.282630      -0.740433  -53.912136       0.741380  -59.004217       2.759183  -63.489005       5.426812  -68.566001       8.712068  -71.394853      11.659135  -66.629580      15.362600  -67.632276      16.827507  -66.459524      19.069327  -63.819523      21.446736  -61.672712      23.532143  -57.538431      25.947815  -52.519889      28.691766  -48.606227      30.646295  -45.000447      31.089437  -41.549866      32.139873  -36.605156      32.956277  -32.010080      34.156692  -29.730629      33.756566  -26.158767      33.714080  -25.821513      34.179648  -23.614658      36.173719  -20.896869      36.977645  -17.991994      35.600074  -13.375742      32.581447  -9.554027      28.675497  -7.825604      26.535234  -7.825604      26.535234  -9.094304      23.363132  -9.564002      22.527385  -9.713885      22.217165  -9.948596      20.367878  -10.496531      16.486580  -11.151919      12.666850  -12.350144       8.800367  -15.446347       4.993373  -20.366139       1.132118  -24.784805      -0.927448  -31.532135      -1.910227  -39.366295      -1.633460

To clarify: the quantity of interest is the projection of the true 3d centroid(i.e. 3d center-of-mass, i.e. 3d center-of-area) onto the unit sphere.

Since all you care about is the direction from the origin to the 3d centroid,you don't need to bother with areas at all;it's easier to just compute the moment (i.e. 3d centroid times area).The moment of the region to the left of a closed path on the unit sphereis half the integral of the leftward unit vector as you walk around the path.This follows from a non-obvious application of Stokes' theorem; see http://www.owlnet.rice.edu/~fjones/chap13.pdf Problem 13-12.

In particular, for a spherical polygon, the moment is the half the sum of(a x b) / ||a x b|| * (angle between a and b) for each pair of consecutive vertices a,b.(That's for the region to the left of the path;negate it for the region to the right of the path.)

(And if you really did want the 3d centroid, just compute the area and divide the moment by it. Comparing areas might also be useful in choosing which of the two regions to call "the polygon".)

Here's some code; it's really simple:

#!/usr/bin/pythonimport mathdef plus(a,b): return [x+y for x,y in zip(a,b)]def minus(a,b): return [x-y for x,y in zip(a,b)]def cross(a,b): return [a[1]*b[2]-a[2]*b[1], a[2]*b[0]-a[0]*b[2], a[0]*b[1]-a[1]*b[0]]def dot(a,b): return sum([x*y for x,y in zip(a,b)])def length(v): return math.sqrt(dot(v,v))def normalized(v): l = length(v); return [1,0,0] if l==0 else [x/l for x in v]def addVectorTimesScalar(accumulator, vector, scalar):    for i in xrange(len(accumulator)): accumulator[i] += vector[i] * scalardef angleBetweenUnitVectors(a,b):    # http://www.plunk.org/~hatch/rightway.php    if dot(a,b) < 0:        return math.pi - 2*math.asin(length(plus(a,b))/2.)    else:        return 2*math.asin(length(minus(a,b))/2.)def sphericalPolygonMoment(verts):    moment = [0.,0.,0.]    for i in xrange(len(verts)):        a = verts[i]        b = verts[(i+1)%len(verts)]        addVectorTimesScalar(moment, normalized(cross(a,b)),                                     angleBetweenUnitVectors(a,b) / 2.)    return momentif __name__ == '__main__':    import sys    def lonlat_degrees_to_xyz(lon_degrees,lat_degrees):        lon = lon_degrees*(math.pi/180)        lat = lat_degrees*(math.pi/180)        coslat = math.cos(lat)        return [coslat*math.cos(lon), coslat*math.sin(lon), math.sin(lat)]    verts = [lonlat_degrees_to_xyz(*[float(v) for v in line.split()])             for line in sys.stdin.readlines()]    #print "verts = "+`verts`    moment = sphericalPolygonMoment(verts)    print "moment = "+`moment`    print "centroid unit direction = "+`normalized(moment)`

For the example polygon, this gives the answer (unit vector):

[-0.7644875430808217, 0.579935445918147, -0.2814847687566214]

This is roughly the same as, but more accurate than, the answer computed by @KobeJohn's code, which uses rough tolerances and planar approximations to the sub-centroids:

[0.7628095787179151, -0.5977153368303585, 0.24669398601094406]

The directions of the two answers are roughly opposite (so I guess KobeJohn's codedecided to take the region to the right of the path in this case).

I think a good approximation would be to compute the center of mass using weighted cartesian coordinates and projecting the result onto the sphere (supposing the origin of coordinates is (0, 0, 0)^T).

Let be (p[0], p[1], ... p[n-1]) the n points of the polygon. The approximative (cartesian) centroid can be computed by:

c = 1 / w * (sum of w[i] * p[i])

whereas w is the sum of all weights and whereas p[i] is a polygon point and w[i] is a weight for that point, e.g.

w[i] = |p[i] - p[(i - 1 + n) % n]| / 2 + |p[i] - p[(i + 1) % n]| / 2

whereas |x| is the length of a vector x. I.e. a point is weighted with half the length to the previous and half the length to the next polygon point.

This centroid c can now projected onto the sphere by:

c' = r * c / |c| 

whereas r is the radius of the sphere.

To consider orientation of polygon (ccw, cw) the result may be

c' = - r * c / |c|.