I want to rotate my volumetric CT data using the euler angles x, y and z. For this, I use SimpleITK. I have read the question from Dr. Jessop - simpleitk-rotation-of-volumetric-data-e-g-mri and I think I have the same problem that my orientation/direction is not an identity matrix. The direction namely is:
0.08716564279125966, 0.0, -0.9961938319005929, 0.9961938319005927, 6.633444000000004e-17, 0.08716564279125968, 0.0, -1.0, 6.12303124808918e-17
However, the solution Dr. Jessop has found is by using an axis-angle orientation so that he can rotate around the z axis only. I want to rotate around all axes by using Euler angles. How can I achieve this?
P.S. I would have commented on Dr.Jessops question to ask it, but I don't have enough reputation points for this.
The code of dr. Jessop:
# This function is from https://github.com/rock-learning/pytransform3d/blob/7589e083a50597a75b12d745ebacaa7cc056cfbd/pytransform3d/rotations.py#L302
def matrix_from_axis_angle(a):
""" Compute rotation matrix from axis-angle.
This is called exponential map or Rodrigues' formula.
Parameters
----------
a : array-like, shape (4,)
Axis of rotation and rotation angle: (x, y, z, angle)
Returns
-------
R : array-like, shape (3, 3)
Rotation matrix
"""
ux, uy, uz, theta = a
c = np.cos(theta)
s = np.sin(theta)
ci = 1.0 - c
R = np.array([[ci * ux * ux + c,
ci * ux * uy - uz * s,
ci * ux * uz + uy * s],
[ci * uy * ux + uz * s,
ci * uy * uy + c,
ci * uy * uz - ux * s],
[ci * uz * ux - uy * s,
ci * uz * uy + ux * s,
ci * uz * uz + c],
])
# This is equivalent to
# R = (np.eye(3) * np.cos(theta) +
# (1.0 - np.cos(theta)) * a[:3, np.newaxis].dot(a[np.newaxis, :3]) +
# cross_product_matrix(a[:3]) * np.sin(theta))
return R
def resample(image, transform):
"""
This function resamples (updates) an image using a specified transform
:param image: The sitk image we are trying to transform
:param transform: An sitk transform (ex. resizing, rotation, etc.
:return: The transformed sitk image
"""
reference_image = image
interpolator = sitk.sitkLinear
default_value = 0
return sitk.Resample(image, reference_image, transform,
interpolator, default_value)
def get_center(img):
"""
This function returns the physical center point of a 3d sitk image
:param img: The sitk image we are trying to find the center of
:return: The physical center point of the image
"""
width, height, depth = img.GetSize()
return img.TransformIndexToPhysicalPoint((int(np.ceil(width/2)),
int(np.ceil(height/2)),
int(np.ceil(depth/2))))
def rotation3d(image, theta_z, show=False):
"""
This function rotates an image across each of the x, y, z axes by theta_x, theta_y, and
theta_z degrees
respectively
:param image: An sitk MRI image
:param theta_x: The amount of degrees the user wants the image rotated around the x axis
:param theta_y: The amount of degrees the user wants the image rotated around the y axis
:param theta_z: The amount of degrees the user wants the image rotated around the z axis
:param show: Boolean, whether or not the user wants to see the result of the rotation
:return: The rotated image
"""
theta_z = np.deg2rad(theta_z)
euler_transform = sitk.Euler3DTransform()
print(euler_transform.GetMatrix())
image_center = get_center(image)
euler_transform.SetCenter(image_center)
direction = image.GetDirection()
print(direction)
axis_angle = (direction[2], direction[5], direction[8], theta_z)
np_rot_mat = matrix_from_axis_angle(axis_angle)
euler_transform.SetMatrix(np_rot_mat.flatten().tolist())
resampled_image = resample(image, euler_transform)
if show:
slice_num = int(input("Enter the index of the slice you would like to see"))
plt.imshow(sitk.GetArrayFromImage(resampled_image)[slice_num])
plt.show()
return resampled_image
To get the rotationmatrix from the euler angle method, this code could be used:
def matrix_from_euler_xyz(e):
"""Compute rotation matrix from xyz Euler angles.
Intrinsic rotations are used to create the transformation matrix
from three concatenated rotations.
The xyz convention is usually used in physics and chemistry.
Parameters
----------
e : array-like, shape (3,)
Angles for rotation around x-, y'-, and z''-axes (intrinsic rotations)
Returns
-------
R : array-like, shape (3, 3)
Rotation matrix
"""
alpha, beta, gamma = e
# We use intrinsic rotations
Qx = matrix_from_angle(0, alpha)
Qy = matrix_from_angle(1, beta)
Qz = matrix_from_angle(2, gamma)
R = Qx.dot(Qy).dot(Qz)
return R
However, the orientation should still be incorporated. Does anyone know how to do this?
This can be performed using SimpleITK using VersorRigid3DTransform and passing it into the resampleitkfilter, which I believe is what this code is doing in a way.
I would suggest using scipy.transform.rotation module to pass in a versor into the rigid3Dtransform then set your direction from the input to output accordingly.
I 'think' this will work even with your off-unit directions but as always with 3D affine transforms this might be missing something that brighter mathematicians can critique:
r = R.from_euler('zyx', [90, 45, 30], degrees=True)
orientation = r.as_quat()
transform = sitk.VersorRigid3DTransform()
translation = sitk.TranslationTransform(3, (x,y,z))
transform.SetTranslation(translation.GetOffset())
#get rot_center from your center of image function as tuple (x,y,z)
rotation = sitk.VersorTransform([orientation[0], orientation[1], orientation[2], orientation[3]],rot_center)
transform.SetRotation(rotation.GetVersor())
transform.SetCenter(rotation.GetCenter())
size = img.GetSize()
bounds = list()
for x in [0, size[0]]:
for y in [0, size[1]]:
for z in [0, size[2]]:
bounds.append(img.TransformIndexToPhysicalPoint((x, y, z)))
# get the physical position of the bounds given transform
trans_bounds = list()
for b in bounds:
trans_bounds.append(transform.TransformPoint((b[0], b[1], b[2])))
trans_bounds = np.array(trans_bounds)
xmin = np.min(trans_bounds.T[0])
xmax = np.max(trans_bounds.T[0])
ymin = np.min(trans_bounds.T[1])
ymax = np.max(trans_bounds.T[1])
zmin = np.min(trans_bounds.T[2])
zmax = np.max(trans_bounds.T[2])
output_origin = (xmin, ymin, zmin)
res_all = (0.5, 0.5, 0.5)
res = res_all[0]
output_size = np.array(
[int(np.round((xmax - xmin) / res)), int(np.round((ymax - ymin) / res)), int(np.round((zmax - zmin) / res))])
output_size = output_size.astype(int)
resampleFilter = sitk.ResampleImageFilter()
resampleFilter.SetTransform(transform.GetInverse())
resampleFilter.SetInterpolator(sitk.sitkLinear)
resampleFilter.SetSize(output_size.tolist())
resampleFilter.SetOutputOrigin(output_origin)
resampleFilter.SetOutputSpacing(res_all)
resampleFilter.SetOutputDirection(img.GetDirection())
resampleFilter.SetDefaultPixelValue(0.0)
I solved it (with the help of zivy from the itk discourse
My answer is this:
# The function which is used to rotate (and make the 3D image isotropic) using euler angles
def rotation3d(image, theta_x, theta_y, theta_z, output_spacing = None, background_value=0.0):
"""
This function rotates an image across each of the x, y, z axes by theta_x, theta_y, and theta_z degrees
respectively (euler ZXY orientation) and resamples it to be isotropic.
:param image: An sitk 3D image
:param theta_x: The amount of degrees the user wants the image rotated around the x axis
:param theta_y: The amount of degrees the user wants the image rotated around the y axis
:param theta_z: The amount of degrees the user wants the image rotated around the z axis
:param output_spacing: Scalar denoting the isotropic output image spacing. If None, then use the smallest
spacing from original image.
:return: The rotated image
"""
euler_transform = sitk.Euler3DTransform (image.TransformContinuousIndexToPhysicalPoint([(sz-1)/2.0 for sz in image.GetSize()]),
np.deg2rad(theta_x),
np.deg2rad(theta_y),
np.deg2rad(theta_z))
# compute the resampling grid for the transformed image
max_indexes = [sz-1 for sz in image.GetSize()]
extreme_indexes = list(itertools.product(*(list(zip([0]*image.GetDimension(),max_indexes)))))
extreme_points_transformed = [euler_transform.TransformPoint(image.TransformContinuousIndexToPhysicalPoint(p)) for p in extreme_indexes]
output_min_coordinates = np.min(extreme_points_transformed, axis=0)
output_max_coordinates = np.max(extreme_points_transformed, axis=0)
# isotropic ouput spacing
if output_spacing is None:
output_spacing = min(image.GetSpacing())
output_spacing = [output_spacing]*image.GetDimension()
output_origin = output_min_coordinates
output_size = [int(((omx-omn)/ospc)+0.5) for ospc, omn, omx in zip(output_spacing, output_min_coordinates, output_max_coordinates)]
output_direction = [1,0,0,0,1,0,0,0,1]
output_pixeltype = image.GetPixelIDValue()
return sitk.Resample(image,
output_size,
euler_transform.GetInverse(),
sitk.sitkLinear,
output_origin,
output_spacing,
output_direction,
background_value,
output_pixeltype)
I am not that experienced in python but improving it thanks to this community! I desperately need a function which takes the input and gives the ouput below:
Input:
1- Latitude/longitude coordinates of the center of circle 1 (e.g. (50.851295, 5.667969) )
2- The radius of circle 1 in meters (e.g. 200)
3- Latitude/longitude coordinates of the center of circle 2 (e.g. (50.844101, 5.725889) )
4- The radius of circle 2 in meters (e.g. 300)
Output: Possible output examples can be;
The intersection points are (50.848295, 5.707969) and (50.849295, 5.717969)
The circles are overlapping
The circles are tangential and the intersection point is (50.847295, 5.705969)
The circles do not intersect
I have examined the similar topics in this platform, other platforms, libraries, tried to combine different solutions but couldn't succeed. Any help is much appreciated!
EDIT:
The problem is solved many thanks to Ture Pålsson who commented below and directed me to whuber's brilliant work in this link https://gis.stackexchange.com/questions/48937/calculating-intersection-of-two-circles Based on that work, I wrote the code below and as far as I tested it works. I want to share it here in case someone might find it helpful. Any feedback is appreciated.
'''
FINDING THE INTERSECTION COORDINATES (LAT/LON) OF TWO CIRCLES (GIVEN THE COORDINATES OF THE CENTER AND THE RADII)
Many thanks to Ture Pålsson who directed me to the right source, the code below is based on whuber's brilliant logic and
explanation here https://gis.stackexchange.com/questions/48937/calculating-intersection-of-two-circles
The idea is that;
1. The points in question are the mutual intersections of three spheres: a sphere centered beneath location x1 (on the
earth's surface) of a given radius, a sphere centered beneath location x2 (on the earth's surface) of a given radius, and
the earth itself, which is a sphere centered at O = (0,0,0) of a given radius.
2. The intersection of each of the first two spheres with the earth's surface is a circle, which defines two planes.
The mutual intersections of all three spheres therefore lies on the intersection of those two planes: a line.
Consequently, the problem is reduced to intersecting a line with a sphere.
Note that "Decimal" is used to have higher precision which is important if the distance between two points are a few
meters.
'''
from decimal import Decimal
from math import cos, sin, sqrt
import math
import numpy as np
def intersection(p1, r1_meter, p2, r2_meter):
# p1 = Coordinates of Point 1: latitude, longitude. This serves as the center of circle 1. Ex: (36.110174, -90.953524)
# r1_meter = Radius of circle 1 in meters
# p2 = Coordinates of Point 2: latitude, longitude. This serves as the center of circle 1. Ex: (36.110174, -90.953524)
# r2_meter = Radius of circle 2 in meters
'''
1. Convert (lat, lon) to (x,y,z) geocentric coordinates.
As usual, because we may choose units of measurement in which the earth has a unit radius
'''
x_p1 = Decimal(cos(math.radians(p1[1]))*cos(math.radians(p1[0]))) # x = cos(lon)*cos(lat)
y_p1 = Decimal(sin(math.radians(p1[1]))*cos(math.radians(p1[0]))) # y = sin(lon)*cos(lat)
z_p1 = Decimal(sin(math.radians(p1[0]))) # z = sin(lat)
x1 = (x_p1, y_p1, z_p1)
x_p2 = Decimal(cos(math.radians(p2[1]))*cos(math.radians(p2[0]))) # x = cos(lon)*cos(lat)
y_p2 = Decimal(sin(math.radians(p2[1]))*cos(math.radians(p2[0]))) # y = sin(lon)*cos(lat)
z_p2 = Decimal(sin(math.radians(p2[0]))) # z = sin(lat)
x2 = (x_p2, y_p2, z_p2)
'''
2. Convert the radii r1 and r2 (which are measured along the sphere) to angles along the sphere.
By definition, one nautical mile (NM) is 1/60 degree of arc (which is pi/180 * 1/60 = 0.0002908888 radians).
'''
r1 = Decimal(math.radians((r1_meter/1852) / 60)) # r1_meter/1852 converts meter to Nautical mile.
r2 = Decimal(math.radians((r2_meter/1852) / 60))
'''
3. The geodesic circle of radius r1 around x1 is the intersection of the earth's surface with an Euclidean sphere
of radius sin(r1) centered at cos(r1)*x1.
4. The plane determined by the intersection of the sphere of radius sin(r1) around cos(r1)*x1 and the earth's surface
is perpendicular to x1 and passes through the point cos(r1)x1, whence its equation is x.x1 = cos(r1)
(the "." represents the usual dot product); likewise for the other plane. There will be a unique point x0 on the
intersection of those two planes that is a linear combination of x1 and x2. Writing x0 = ax1 + b*x2 the two planar
equations are;
cos(r1) = x.x1 = (a*x1 + b*x2).x1 = a + b*(x2.x1)
cos(r2) = x.x2 = (a*x1 + b*x2).x2 = a*(x1.x2) + b
Using the fact that x2.x1 = x1.x2, which I shall write as q, the solution (if it exists) is given by
a = (cos(r1) - cos(r2)*q) / (1 - q^2),
b = (cos(r2) - cos(r1)*q) / (1 - q^2).
'''
q = Decimal(np.dot(x1, x2))
if q**2 != 1 :
a = (Decimal(cos(r1)) - Decimal(cos(r2))*q) / (1 - q**2)
b = (Decimal(cos(r2)) - Decimal(cos(r1))*q) / (1 - q**2)
'''
5. Now all other points on the line of intersection of the two planes differ from x0 by some multiple of a vector
n which is mutually perpendicular to both planes. The cross product n = x1~Cross~x2 does the job provided n is
nonzero: once again, this means that x1 and x2 are neither coincident nor diametrically opposite. (We need to
take care to compute the cross product with high precision, because it involves subtractions with a lot of
cancellation when x1 and x2 are close to each other.)
'''
n = np.cross(x1, x2)
'''
6. Therefore, we seek up to two points of the form x0 + t*n which lie on the earth's surface: that is, their length
equals 1. Equivalently, their squared length is 1:
1 = squared length = (x0 + t*n).(x0 + t*n) = x0.x0 + 2t*x0.n + t^2*n.n = x0.x0 + t^2*n.n
'''
x0_1 = [a*f for f in x1]
x0_2 = [b*f for f in x2]
x0 = [sum(f) for f in zip(x0_1, x0_2)]
'''
The term with x0.n disappears because x0 (being a linear combination of x1 and x2) is perpendicular to n.
The two solutions easily are t = sqrt((1 - x0.x0)/n.n) and its negative. Once again high precision
is called for, because when x1 and x2 are close, x0.x0 is very close to 1, leading to some loss of
floating point precision.
'''
if (np.dot(x0, x0) <= 1) & (np.dot(n,n) != 0): # This is to secure that (1 - np.dot(x0, x0)) / np.dot(n,n) > 0
t = Decimal(sqrt((1 - np.dot(x0, x0)) / np.dot(n,n)))
t1 = t
t2 = -t
i1 = x0 + t1*n
i2 = x0 + t2*n
'''
7. Finally, we may convert these solutions back to (lat, lon) by converting geocentric (x,y,z) to geographic
coordinates. For the longitude, use the generalized arctangent returning values in the range -180 to 180
degrees (in computing applications, this function takes both x and y as arguments rather than just the
ratio y/x; it is sometimes called "ATan2").
'''
i1_lat = math.degrees( math.asin(i1[2]))
i1_lon = math.degrees( math.atan2(i1[1], i1[0] ) )
ip1 = (i1_lat, i1_lon)
i2_lat = math.degrees( math.asin(i2[2]))
i2_lon = math.degrees( math.atan2(i2[1], i2[0] ) )
ip2 = (i2_lat, i2_lon)
return [ip1, ip2]
elif (np.dot(n,n) == 0):
return("The centers of the circles can be neither the same point nor antipodal points.")
else:
return("The circles do not intersect")
else:
return("The centers of the circles can be neither the same point nor antipodal points.")
'''
Example: the output of below is [(36.989311051533505, -88.15142628069133), (38.2383796094578, -92.39048549120287)]
intersection_points = intersection((37.673442, -90.234036), 107.5*1852, (36.109997, -90.953669), 145*1852)
print(intersection_points)
'''
Depending on the precision you need, you may or may not consider the Earth as a sphere. In the second case, calculations become more complex.
The best option for precise measurements when the radius is small (as in your example) is to use a projection (UTM for example) and then apply the common flat euclidean calculations.
Let's first copy the flat circle intersection function from https://stackoverflow.com/a/55817881/2148416:
def circle_intersection(x0, y0, r0, x1, y1, r1):
d = math.sqrt((x1 - x0) ** 2 + (y1 - y0) ** 2)
if d > r0 + r1: # non intersecting
return None
if d < abs(r0 - r1): # one circle within other
return None
if d == 0 and r0 == r1: # coincident circles
return None
a = (r0 ** 2 - r1 ** 2 + d ** 2) / (2 * d)
h = math.sqrt(r0 ** 2 - a ** 2)
x2 = x0 + a * (x1 - x0) / d
y2 = y0 + a * (y1 - y0) / d
x3 = x2 + h * (y1 - y0) / d
y3 = y2 - h * (x1 - x0) / d
x4 = x2 - h * (y1 - y0) / d
y4 = y2 + h * (x1 - x0) / d
return (x3, y3), (x4, y4)
The precise calculation for a small radius (up to a few kilometers) can be done in UTM coordinates with the help of the utm library. It handles all the complications regarding the fact the the Earth is more an ellipsoid than a sphere:
import utm
def geo_circle_intersection(latlon0, radius0, latlon1, radius1):
# Convert lat/lon to UTM
x0, y0, zone, letter = utm.from_latlon(latlon0[0], latlon0[1])
x1, y1, _, _ = utm.from_latlon(latlon1[0], latlon1 [1], force_zone_number=zone)
# Calculate intersections in UTM coordinates
a_utm, b_utm = circle_intersection(x0, y0, r0, x1, y1, r1)
# Convert intersections from UTM back to lat/lon
a = utm.to_latlon(a_utm[0], a_utm[1], zone, letter)
b = utm.to_latlon(b_utm[0], b_utm[1], zone, letter)
return a, b
Using your example (with slightly larger radii):
>>> p0 = 50.851295, 5.667969
>>> r0 = 2000
>>> p1 = 50.844101, 5.725889
>>> r1 = 3000
>>> a, b = geo_circle_intersection(p0, r0, p1, r1)
>>> print(a)
(50.836848562566004, 5.684869539768468)
>>> print(b)
(50.860635308778285, 5.692236858407678)
I have a midpoint (x,y) and i need to create a square polygon with random orientation using a 2D (random) planar rotation.
def get_square_plot(x, y, side):
return [(x-(side/2), y+(side/2)), (x+(side/2), y+(side/2)), (x+(side/2), y-(side/2)), (x-(side/2), y-(side/2))]
This function creates the vertices of a square polygon without a specific orientation. I wish to improve this function adding the possibility to rotation randomly these vertices (and with a specific angle if is possible)
If I've understood you correctly, this should be able to do what you want:
from math import sin, cos, radians
def rotated_square(cx, cy, size, degrees=0):
""" Calculate coordinates of a rotated square centered at 'cx, cy'
given its 'size' and rotation by 'degrees' about its center.
"""
h = size/2
l, r, b, t = cx-h, cx+h, cy-h, cy+h
a = radians(degrees)
cosa, sina = cos(a), sin(a)
pts = [(l, b), (l, t), (r, t), (r, b)]
return [(( (x-cx)*cosa + (y-cy)*sina) + cx,
(-(x-cx)*sina + (y-cy)*cosa) + cy) for x, y in pts]
print rotated_square(50, 50, 100)
Output:
[(0.0, 0.0), (0.0, 100.0), (100.0, 100.0), (100.0, 0.0)]
Note that in the general case, the resulting coordinates won't be integers.
What this does effectively is first translate each coordinate to the origin by subtracting cx,cy from it, rotates that by the angle, and then un-translates it back by the same amount. This is necessary to compensate for the fact that rotation formulas usually are relative to origin of a coordinate system.
having determined the four corner coordinates, you can rotate them relative to the origin (or midpoint) using a simple 2D Matrix Rotation:
http://en.wikipedia.org/wiki/Rotation_%28mathematics%29 (search for 2D rotation equation)
x' = x cos(theta) - y sin(theta)
y' = x sin(theta) + y cos(theta)
You can use the built-in Python Math library for the cos/sin functions: http://docs.python.org/2/library/math.html section 9.2.3
Math.cos(theta)
Math.sin(theta)
I hope this is of some use!
I'm trying to work out how best to locate the centroid of an arbitrary shape draped over a unit sphere, with the input being ordered (clockwise or anti-cw) vertices for the shape boundary. The density of vertices is irregular along the boundary, so the arc-lengths between them are not generally equal. Because the shapes may be very large (half a hemisphere) it is generally not possible to simply project the vertices to a plane and use planar methods, as detailed on Wikipedia (sorry I'm not allowed more than 2 hyperlinks as a newcomer). A slightly better approach involves the use of planar geometry manipulated in spherical coordinates, but again, with large polygons this method fails, as nicely illustrated here. On that same page, 'Cffk' highlighted this paper which describes a method for calculating the centroid of spherical triangles. I've tried to implement this method, but without success, and I'm hoping someone can spot the problem?
I have kept the variable definitions similar to those in the paper to make it easier to compare. The input (data) is a list of longitude/latitude coordinates, converted to [x,y,z] coordinates by the code. For each of the triangles I have arbitrarily fixed one point to be the +z-pole, the other two vertices being composed of a pair of neighboring points along the polygon boundary. The code steps along the boundary (starting at an arbitrary point), using each boundary segment of the polygon as a triangle side in turn. A sub-centroid is determined for each of these individual spherical triangles and they are weighted according to triangle area and added to calculate the total polygon centroid. I don't get any errors when running the code, but the total centroids returned are clearly wrong (I have run some very basic shapes where the centroid location is unambiguous). I haven't found any sensible pattern in the location of the centroids returned...so at the moment I'm not sure what is going wrong, either in the math or code (although, the suspicion is the math).
The code below should work copy-paste as is if you would like to try it. If you have matplotlib and numpy installed, it will plot the results (it will ignore plotting if you don't). You just have to put the longitude/latitude data below the code into a text file called example.txt.
from math import *
try:
import matplotlib as mpl
import matplotlib.pyplot
from mpl_toolkits.mplot3d import Axes3D
import numpy
plotting_enabled = True
except ImportError:
plotting_enabled = False
def sph_car(point):
if len(point) == 2:
point.append(1.0)
rlon = radians(float(point[0]))
rlat = radians(float(point[1]))
x = cos(rlat) * cos(rlon) * point[2]
y = cos(rlat) * sin(rlon) * point[2]
z = sin(rlat) * point[2]
return [x, y, z]
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 dot
def plot(poly_xyz, g_xyz):
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(*poly_xyz)
ax.plot(x, y, z)
ax.plot(x, y, zs=0)
# plot the alleged 3d and flattened centroid
x, y, z = g_xyz
ax.scatter(x, y, z, c='r')
ax.scatter(x, y, 0, c='r')
# display
ax.set_xlim3d(-1, 1)
ax.set_ylim3d(-1, 1)
ax.set_zlim3d(0, 1)
mpl.pyplot.show()
lons, lats, v = list(), list(), list()
# put the two-column data at the bottom of the question into a file called
# example.txt in the same directory as this script
with open('example.txt') as f:
for line in f.readlines():
sep = line.split()
lons.append(float(sep[0]))
lats.append(float(sep[1]))
# convert spherical coordinates to cartesian
for lon, lat in zip(lons, lats):
v.append(sph_car([lon, lat, 1.0]))
# z unit vector/pole ('north pole'). This is an arbitrary point selected to act as one
#(fixed) vertex of the summed spherical triangles. The other two vertices of any
#triangle are composed of neighboring vertices from the polygon boundary.
np = [0.0, 0.0, 1.0]
# Gx,Gy,Gz are the cartesian coordinates of the calculated centroid
Gx, Gy, Gz = 0.0, 0.0, 0.0
for i in range(-1, len(v) - 1):
# cycle through the boundary vertices of the polygon, from 0 to n
if all((v[i][0] != v[i+1][0],
v[i][1] != v[i+1][1],
v[i][2] != v[i+1][2])):
# this just ignores redundant points which are common in my larger input files
# A,B,C are the internal angles in the triangle: 'np-v[i]-v[i+1]-np'
A = asin(sqrt((dprod(np, xprod(v[i], v[i+1])))**2
/ ((1 - (dprod(v[i+1], np))**2) * (1 - (dprod(np, v[i]))**2))))
B = asin(sqrt((dprod(v[i], xprod(v[i+1], np)))**2
/ ((1 - (dprod(np , v[i]))**2) * (1 - (dprod(v[i], v[i+1]))**2))))
C = asin(sqrt((dprod(v[i + 1], xprod(np, v[i])))**2
/ ((1 - (dprod(v[i], v[i+1]))**2) * (1 - (dprod(v[i+1], np))**2))))
# A/B/Cbar are the vertex angles, such that if 'O' is the sphere center, Abar
# is the angle (v[i]-O-v[i+1])
Abar = acos(dprod(v[i], v[i+1]))
Bbar = acos(dprod(v[i+1], np))
Cbar = acos(dprod(np, v[i]))
# e is the 'spherical excess', as defined on wikipedia
e = A + B + C - pi
# mag1/2/3 are the magnitudes of vectors np,v[i] and v[i+1].
mag1 = 1.0
mag2 = float(sqrt(v[i][0]**2 + v[i][1]**2 + v[i][2]**2))
mag3 = float(sqrt(v[i+1][0]**2 + v[i+1][1]**2 + v[i+1][2]**2))
# vec1/2/3 are cross products, defined here to simplify the equation below.
vec1 = xprod(np, v[i])
vec2 = xprod(v[i], v[i+1])
vec3 = xprod(v[i+1], np)
# multiplying vec1/2/3 by e and respective internal angles, according to the
#posted paper
for x in range(3):
vec1[x] *= Cbar / (2 * e * mag1 * mag2
* sqrt(1 - (dprod(np, v[i])**2)))
vec2[x] *= Abar / (2 * e * mag2 * mag3
* sqrt(1 - (dprod(v[i], v[i+1])**2)))
vec3[x] *= Bbar / (2 * e * mag3 * mag1
* sqrt(1 - (dprod(v[i+1], np)**2)))
Gx += vec1[0] + vec2[0] + vec3[0]
Gy += vec1[1] + vec2[1] + vec3[1]
Gz += vec1[2] + vec2[2] + vec3[2]
approx_expected_Gxyz = (0.78, -0.56, 0.27)
print('Approximate Expected Gxyz: {0}\n'
' Actual Gxyz: {1}'
''.format(approx_expected_Gxyz, (Gx, Gy, Gz)))
if plotting_enabled:
plot(v, (Gx, Gy, Gz))
Thanks in advance for any suggestions or insight.
EDIT: Here is a figure that shows a projection of the unit sphere with a polygon and the resulting centroid I calculate from the code. Clearly, the centroid is wrong as the polygon is rather small and convex but yet the centroid falls outside its perimeter.
EDIT: Here is a highly-similar set of coordinates to those above, but in the original [lon,lat] format I normally use (which is now converted to [x,y,z] by the updated code).
-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
EDIT: A couple more examples...with 4 vertices defining a perfect square centered at [1,0,0] I get the expected result:
However, from a non-symmetric triangle I get a centroid that is nowhere close...the centroid actually falls on the far side of the sphere (here projected onto the front side as the antipode):
Interestingly, the centroid estimation appears 'stable' in the sense that if I invert the list (go from clockwise to counterclockwise order or vice-versa) the centroid correspondingly inverts exactly.
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 = True
except ImportError:
plotting_enabled = False
def 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 cartesians
def 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, z
def 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 area
def 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 angles
def 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 dot
def 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 top
main()
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 sphere
is 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 Frank Jones's vector calculus book, chapter 13 Problem 13-12.
In particular, for a spherical polygon, the moment is 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/python
import math
def 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] * scalar
def angleBetweenUnitVectors(a,b):
# https://www.plunk.org/~hatch/rightway.html
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 moment
if __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 code
decided 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|.
Sorry I (as a newly registered user) had to write a new post instead of just voting/commenting on the above answer by Don Hatch. Don's answer, I think, is the best and most elegant. It is mathematically rigorous in computing the center of mass (first moment of mass) in a simple way when applying to the spherical polygon.
Kobe John's answer is a good approximation but only satisfactory for smaller areas. I also noticed a few glitches in the code. Firstly, the reference point should be projected to the spherical surface to compute the actual spherical area. Secondly, function points_are_equivalent() might need to be refined to avoid divided-by-zero.
The approximation error in Kobe's method lies in the calculation of the centroid of spherical triangles. The sub-centroid is NOT the center of mass of the spherical triangle but the planar one. This is not an issue if one is to determine that single triangle (sign may flip, see below). It is also not an issue if triangles are small (e.g. a dense triangulation of the polygon).
A few simple tests could illustrate the approximation error. For example if we use just four points:
10 -20
10 20
-10 20
-10 -20
The exact answer is (1,0,0) and both methods are good. But if you throw in a few more points along one edge (e.g. add {10,-15},{10,-10}... to the first edge), you'll see the results from Kobe's method start to shift. Further more, if you increase the longitude from [10,-10] to [100,-100], you'll see Kobe's result flips the direction. A possible improvement might be to add another level(s) for sub-centroid calculation (basically refine/reduce sizes of triangles).
For our application, the spherical area boundary is composed of multiple arcs and thus not polygon (i.e. the arc is not part of great circle). But this will just be a little more work to find the n-vector in the curve integration.
EDIT: Replacing the subcentroid calculation with the one given in Brock's paper should fix Kobe's method. But I did not try though.