Python: find area of polygon from xyz coordinates

Jamie Bull picture Jamie Bull · Sep 28, 2012 · Viewed 21.3k times · Source

I'm trying to use the shapely.geometry.Polygon module to find the area of polygons but it performs all calculations on the xy plane. This is fine for some of my polygons but others have a z dimension too so it's not quite doing what I'd like.

Is there a package which will either give me the area of a planar polygon from xyz coordinates, or alternatively a package or algorithm to rotate the polygon to the xy plane so that i can use shapely.geometry.Polygon().area?

The polygons are represented as a list of tuples in the form [(x1,y1,z1),(x2,y2,z3),...(xn,yn,zn)].

Answer

Tom Smilack picture Tom Smilack · Sep 28, 2012

Here is the derivation of a formula for calculating the area of a 3D planar polygon

Here is Python code that implements it:

#determinant of matrix a
def det(a):
    return a[0][0]*a[1][1]*a[2][2] + a[0][1]*a[1][2]*a[2][0] + a[0][2]*a[1][0]*a[2][1] - a[0][2]*a[1][1]*a[2][0] - a[0][1]*a[1][0]*a[2][2] - a[0][0]*a[1][2]*a[2][1]

#unit normal vector of plane defined by points a, b, and c
def unit_normal(a, b, c):
    x = det([[1,a[1],a[2]],
             [1,b[1],b[2]],
             [1,c[1],c[2]]])
    y = det([[a[0],1,a[2]],
             [b[0],1,b[2]],
             [c[0],1,c[2]]])
    z = det([[a[0],a[1],1],
             [b[0],b[1],1],
             [c[0],c[1],1]])
    magnitude = (x**2 + y**2 + z**2)**.5
    return (x/magnitude, y/magnitude, z/magnitude)

#dot product of vectors a and b
def dot(a, b):
    return a[0]*b[0] + a[1]*b[1] + a[2]*b[2]

#cross product of vectors a and b
def cross(a, b):
    x = a[1] * b[2] - a[2] * b[1]
    y = a[2] * b[0] - a[0] * b[2]
    z = a[0] * b[1] - a[1] * b[0]
    return (x, y, z)

#area of polygon poly
def area(poly):
    if len(poly) < 3: # not a plane - no area
        return 0

    total = [0, 0, 0]
    for i in range(len(poly)):
        vi1 = poly[i]
        if i is len(poly)-1:
            vi2 = poly[0]
        else:
            vi2 = poly[i+1]
        prod = cross(vi1, vi2)
        total[0] += prod[0]
        total[1] += prod[1]
        total[2] += prod[2]
    result = dot(total, unit_normal(poly[0], poly[1], poly[2]))
    return abs(result/2)

And to test it, here's a 10x5 square that leans over:

>>> poly = [[0, 0, 0], [10, 0, 0], [10, 3, 4], [0, 3, 4]]
>>> poly_translated = [[0+5, 0+5, 0+5], [10+5, 0+5, 0+5], [10+5, 3+5, 4+5], [0+5, 3+5, 4+5]]
>>> area(poly)
50.0
>>> area(poly_translated)
50.0
>>> area([[0,0,0],[1,1,1]])
0

The problem originally was that I had oversimplified. It needs to calculate the unit vector normal to the plane. The area is half of the dot product of that and the total of all the cross products, not half of the sum of all the magnitudes of the cross products.

This can be cleaned up a bit (matrix and vector classes would make it nicer, if you have them, or standard implementations of determinant/cross product/dot product), but it should be conceptually sound.