How would I get the points of a sphere with the radius r? Im looking for something like bresenham's circle drawing algorithm but in 3d. I tried just using the 2d algorithm and drawing circles in increasing sizes behind each other, but that doesnt quite work.
(The picture is distorted around the edges. Thats normal.)
Do you know of anything that could fix that? I could draw a nother smaler circle inside of the bigger one, but thats cheating. Yes, I went through the existing Questions but couldnt find any that fit my purpose.
Edit:
def drawCircle(self, xyz, r):
xc, yc, zc = xyz
coords = []
def drawC(xc, yc, zc, x, y):
coords.append((xc+x, yc+y, zc))
coords.append((xc-x, yc+y, zc))
coords.append((xc+x, yc-y, zc))
coords.append((xc-x, yc-y, zc))
coords.append((xc+y, yc+x, zc))
coords.append((xc-y, yc+x, zc))
coords.append((xc+y, yc-x, zc))
coords.append((xc-y, yc-x, zc))
x = 0
y = r
d = 3 - 2 * r
drawC(xc, yc, zc, x, y)
while y >= x:
x += 1
if (d > 0):
y -= 1
d = d + 4 * (x - y) + 10
else:
d = d + 4 * x + 6
drawC(xc, yc, zc, x, y)
for c in coords:
self.drawPixel(c)
return coords
This function takes a x,y,z point and a radius, calculates all points on the circle and draws them into a 3d array, witch can be drawn to the screen by another function. Here is an example with r = 10. (The circle is pink. Im just testing color.) Im looking or a version without np if possible.
You can create a centered sphere by using the code below.
import numpy as np
r=2
y,x,z = np.ogrid[-r:r+1, -r:r+1, -r:r+1,]
mask = x**2 + y**2 + z**2 <= r**2
print(mask)
I am trying to create a circle on co-ordinate plane and fill the pixels in it with conditional colors. However, the circle also need to show a small triangular notch at the bottom from the center. Something like the attached picture.
I have used the matplotlib's patches class to create acircle and tried different values in attributes but of no help. I googled enough but I couldn't find it.
circle = matplotlib.patches.Circle((0,0),150,facecolor='lightgrey')
ax.add_patch(circle)
Sample
Can someone please help me or provide me a hint or direct me to right library which can do this.
Your problem looks like 2D vector graphics which in this case you should look for SVG
if it is a CAD problem then you can check some CAD libraries
- Python module for parametric CAD
I think CAD approach is much better for engineering operations which you can find https://www.freecadweb.org/wiki/Part_Slice and https://www.freecadweb.org/wiki/Part_SliceApart#Scripting
or for SVG approach found something: https://inkscape.org/~Moini/%E2%98%85multi-bool-extension-cut-difference-division
If you have to do that in mathplotlib then you can export/import vector graphics into plots
I tried to build the proposed shape using parametric equations.
There are some approximations, but if you get the exact geometry equations you could refine this to a better version.
In fact, the key is to define the correct equations ... the ideal case would be to ensure continuity.
This example has some caveats: approximations when joining the 2 circles due to geometrical simplification when drawing the small circle from pi to 0. Small circle start/end angles should be chosen as the interception of the 2 full circles for a more accurate shape continuity. But again, in the end it depends entirely on your shape specification
Heavly inspired from Plot equation showing a circle
import math
import numpy as np
import matplotlib.pyplot as plt
def compute_x_values(r, theta):
return r * np.cos(theta)
def compute_y_values(r, theta):
return r * np.sin(theta)
def compute_circle(r, theta):
return compute_x_values(r, theta), compute_y_values(r, theta)
def build_big_circle(crop_angle, offset, radius):
start_angle = offset + crop_angle
end_angle = offset + (2 * np.pi) - crop_angle
theta = np.linspace(start_angle, end_angle, 250)
# compute main circle vals
x, y = compute_circle(radius, theta)
return x, y
r = 1
offset = - np.pi / 2
crop_angle = np.pi / 20
x, y = build_big_circle(crop_angle, offset, r)
# now the other form:
# its a half circle from pi to 0
theta2 = np.linspace(np.pi, 0, 100)
# according our code above, angular space left on the circle for the notch is
missing_angle = crop_angle * 2
# the length between to points on a circle is given by the formula
# length = 2 * r * sin(angle/2)
l = math.sin(missing_angle / 2) * r * 2
# we want half the length for a future radius
r2 = l / 2
# the above lines could be optimized to this
# r2 = math.sin(crop_angle) * r
# but I kept intermediate steps for sake of geometric clarity
# equation is same of a circle
x1, y1 = compute_circle(r2, theta2)
# change center on y axis to - big circle radius
y1 = y1 - r
# merge the 2
x_total = np.append(x, x1)
y_total = np.append(y, y1)
# create the global figure
fig, ax = plt.subplots(1)
ax.plot(x_total, y_total)
ax.fill(x_total, y_total, facecolor='lightgrey', linewidth=1)
ax.set_aspect(1)
plt.show()
fig, ax = plt.subplots(1)
ax.plot(x, y)
ax.plot(x1, y1)
ax.set_aspect(1)
plt.show()
I am struggling with layering and zorder in python. I am making a 3D plot using matplotlib with three relevant elements: A surface_plot of a planet, a surface_plot of rings around that planet, and a contourf image that shows the planet's shadow cast onto the rings.
I want the graphics to display exactly how this scenario would look in real life, with the rings going around the planet and the shadow residing across the rings in the appropriate spot. If the shadow is behind the planet for a given POV, I want the shadow to be blocked by the planet, and vice versa if the shadow is in front of the planet for a given POV.
To be clear, this is ONLY a layering issue. I have the planet, rings, and shadow all plotting correctly. However, the shadow will not ever display in front of the planet. It acts as though the planet is "blocking" the shadow, even though the planet is supposed to be underneath the shadow in terms of layering.
I have tried every single thing I can think of in terms of zorder and rearranging which order the various plot elements are called to be drawn. The rings DO correctly display in front of the planet, but the shadow will not.
My actual code is very long. here are the relevant parts:
Plot setup:
def orthographic_proj(zfront, zback):
a = (zfront+zback)/(zfront-zback)
b = -2*(zfront*zback)/(zfront-zback)
return np.array([[1,0,0,0],
[0,1,0,0],
[0,0,a,b],
[0,0,0,zback]])
def setup_saturn_plot(ax3, elev, azim, drawz, drawxy,view):
#ax3.set_aspect('equal','box')
ax3.view_init(elev=elev, azim=azim)
if(view=="top" or view == "Top" or view == "TOP"):
ax3.dist = 5.5
if(view=="star" or view == "Star" or view == "STAR"):
ax3.dist = 5.0 #4.5 is best value
proj3d.persp_transformation = orthographic_proj
# hide grid and background
ax3.w_xaxis.set_pane_color((1.0, 1.0, 1.0, 1.0))
ax3.w_yaxis.set_pane_color((1.0, 1.0, 1.0, 1.0))
ax3.w_zaxis.set_pane_color((1.0, 1.0, 1.0, 1.0))
ax3.grid(False)
# hide z axis in orthographic top view, xy axes in star view
if (drawz == False):
ax3.w_zaxis.line.set_lw(0.)
ax3.set_zticks([])
if (drawz == True):
ax3.set_zlabel('Z (1000 km)',fontsize=12)
if (drawxy == False):
ax3.w_xaxis.line.set_lw(0.)
ax3.set_xticks([])
ax3.w_yaxis.line.set_lw(0.)
ax3.set_yticks([])
if (drawxy == True):
ax3.set_xlabel('X (1000 km)',fontsize=12)
ax3.set_ylabel('Y (1000 km)',fontsize=12)
Planet:
def draw_saturn(ax3, elev, azim):
# Saturn dimensions
radius = 60268. / 1000.
radius_pole = 54364. / 1000.
# draw Saturn
phi, theta = np.mgrid[0.0:np.pi:100j, 0.0:2.0*np.pi:100j]
x = radius*np.sin(phi)*np.cos(theta)
y = radius*np.sin(phi)*np.sin(theta)
z = radius_pole*np.cos(phi)
line3 = ax3.plot_surface(x, y, z, color="w", edgecolor='b', rstride = 8, cstride=5, shade=False, lw=0.25)
#line3 = ax3.plot_wireframe(x, y, z, color="w", edgecolor='b', rstride = 5, cstride=5, lw=0.25)
ax3.tick_params(labelsize=10)
rings:
def draw_rings(ax3, elev, azim, draw_mode):
# Saturn dimensions
radius = 60268. / 1000.
# Saturn rings
dringmin = 1.110 * radius
dringmax = 1.236 * radius
cringmin = 1.239 * radius
titanringlet = 1.292 * radius
maxwellgap = 1.452 * radius
cringmax = 1.526 * radius
bringmin = 1.526 * radius
bringmax = 1.950 * radius
aringmin = 2.030 * radius
enckegap = 2.214 * radius
keelergap = 2.265 * radius
aringmax = 2.270 * radius
fringmin = 2.320 * radius
gringmin = 2.754 * radius
gringmax = 2.874 * radius
eringmin = 2.987 * radius
eringmax = 7.964 * radius
if (draw_mode == 'back'):
offset = -azim*np.pi/180. - 0.5*np.pi
if (draw_mode == 'front'):
offset = -azim*np.pi/180. + 0.5*np.pi
rad, theta = np.mgrid[dringmin:dringmax:4j, 0.0-offset:1.0*np.pi-offset:100j]
x = rad * np.cos(theta)
y = rad * np.sin(theta)
z = 0. * rad
line1 = ax3.plot_surface(x, y, z, color="w", edgecolor='b', rstride = 8, cstride=25, shade=False, lw=0.25,alpha=0.)
rad, theta = np.mgrid[cringmin:cringmax:4j, 0.0-offset:1.0*np.pi-offset:100j]
x = rad * np.cos(theta)
y = rad * np.sin(theta)
z = 0. * rad
line2 = ax3.plot_surface(x, y, z, color="w", edgecolor='b', rstride = 8, cstride=25, shade=False, lw=0.25,alpha=0.)
rad, theta = np.mgrid[bringmin:bringmax:4j, 0.0-offset:1.0*np.pi-offset:100j]
x = rad * np.cos(theta)
y = rad * np.sin(theta)
z = 0. * rad
line3 = ax3.plot_surface(x, y, z, color="w", edgecolor='b', rstride = 8, cstride=25, shade=False, lw=0.25,alpha=0.)
rad, theta = np.mgrid[aringmin:aringmax:4j, 0.0-offset:1.0*np.pi-offset:100j]
x = rad * np.cos(theta)
y = rad * np.sin(theta)
z = 0. * rad
line4 = ax3.plot_surface(x, y, z, color="w", edgecolor='b', rstride = 8, cstride=25, shade=False, lw=0.25,alpha=0.)
rad, theta = np.mgrid[fringmin:1.005*fringmin:2j, 0.0-offset:1.0*np.pi-offset:100j]
x = rad * np.cos(theta)
y = rad * np.sin(theta)
z = 0. * rad
line7 = ax3.plot_surface(x, y, z, color="w", edgecolor='b', rstride = 8, cstride=25, shade=False, lw=0.1,alpha=0.)
Shadow:
def draw_shadowboundary(ax3, sundir):
sqrt = np.sqrt
#azimuthal angle between x direction and direction of sun
alpha = np.arctan2(sundir[1],sundir[0])
#adjustments to keep -pi/2 < alpha < pi/2
alphaadj = 0.*np.pi/180.
if (alpha<0.):
alpha += 2.*np.pi
if ((alpha >= np.pi/2.) & (alpha <= np.pi)):
alpha += np.pi
alphaadj = np.pi
if ((alpha > np.pi) & (alpha <= 3.*np.pi/2.)):
alpha -= np.pi
alphaadj = np.pi
if (alpha>3.*np.pi/2.):
alpha-=2*np.pi
#azimuthal angle between x direction and northern summer -- found using VIMS_2005_14_OMICET and VIMS_2017_053_ALPORI to define eq. of plane of Sun's annual path in chosen coordinate system: -0.193318*x + 0.1963755*y + 0.5471502*z = 0
beta = 44.5505*np.pi/180.
#Saturn's obliquity -- from NASA fact sheet
psi = 26.73*np.pi/180.
#Saturn's oblateness -- from NASA fact sheet
obl = 0.09796
#helpful definitions for optimization
cpsic = np.cos(psi*np.cos(alpha+beta))
spsic = np.sin(psi*np.cos(alpha+beta))
calpha = np.cos(alpha)
salpha = np.sin(alpha)
#Saturn's projected shorter planetary axis as seen by the sun & ring inner edge
req = 60268. / 1000.
b = req*sqrt((1.-obl)*(1.-obl)*cpsic*cpsic + spsic*spsic)
ringstart = 1.239 * req
ringend = 2.270 * req
#shadow boundary of Saturn's rings -- can approximate using a=inf and cancelling terms
a = 9.582*1.496*10.**5
shadowline = lambda x,y : (1/a)*sqrt((req*salpha*(-a+x*calpha*cpsic+y*salpha)*(y*calpha-x*cpsic*salpha)/sqrt((y*calpha-x*cpsic*salpha)**2 + (x*spsic)**2) + calpha*(a*cpsic*(x*calpha*cpsic+y*salpha) + b*x*(a-x*calpha*cpsic-y*salpha)*spsic*spsic/sqrt((y*calpha-x*cpsic*salpha)**2 + (x*spsic)**2)))**2 + (req*calpha*(a-x*calpha*cpsic-y*salpha)*(y*calpha-x*cpsic*salpha)/sqrt((y*calpha-x*cpsic*salpha)**2 + (x*spsic)**2) + salpha*(a*cpsic*(x*calpha*cpsic+y*salpha)+b*x*(a-x*calpha*cpsic-y*salpha)*spsic*spsic/sqrt((y*calpha-x*cpsic*salpha)**2 + (x*spsic)**2)))**2)
#azimuthal radius & antisolar angle for inequalities
radius = lambda x,y : np.sqrt(x**2+y**2)
anti = lambda x,y : abs(np.arctan2(y,x)-(alpha-alphaadj))
#properties of shadow
samples=1200
d = np.linspace(-3*req,3*req,samples)
x,y = np.meshgrid(d,d)
#z = ((radius(x,y)<=shadowline(x,y)) & (ringstart<=radius(x,y)) & (np.pi/2<=anti(x,y)) & (anti(x,y)<=3.*np.pi/2)).astype(int)
z = ((radius(x,y)<=shadowline(x,y)) & (ringstart<=radius(x,y)) & (radius(x,y)<=ringend) & (np.pi/2<=anti(x,y)) & (anti(x,y)<=3.*np.pi/2)).astype(int)
cmap = matplotlib.colors.ListedColormap(["k","k"])
#add shadow to plot
ax3.contourf(x,y,z, [0.5,1.50001], cmap=cmap,alpha=0.5)
Combine graphics:
import matplotlib
import numpy
from math import *
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.mplot3d import Axes3D # <--- This is important for 3d plotting
from mpl_toolkits.mplot3d import proj3d
def plot_results(phi, theta, sundir=[0.5, 0.5]):
#plot_names.append("occultation_track_" + starname)
fig2 = plt.figure(figsize=(9,9))
ax3 = fig2.add_subplot(111, projection='3d')
setup_saturn_plot(ax3, phi, theta, False, False, "star")
draw_saturn(ax3, phi, theta)
draw_rings(ax3, phi, theta, 'back')
draw_rings(ax3, phi, theta, 'front')
draw_shadowboundary(ax3,sundir)
ax3.set_xlim([-200, 200])
ax3.set_ylim([-200, 200])
ax3.set_zlim([-200, 200])
plot_results(phi=40, theta=50, sundir = (30,60))
The code produces an image like this:
The grey shadow is supposed to be residing on the rings in front of the planet. However, it won't display in front of the planet, so only the little sliver of shadow to the right of the planet is actually appearing. The shadow displays correctly in all scenarios except when it needs to go in front the planet.
Any fixes for this?
I'm working on getting my head around this code at the moment, but in the meantime, at least so far, this seams to be a known issue with matplotlib3d.
As #TheImportanceOfBeingErnest pointed out a long time ago, this issue appears in the mpl3d faq
My 3D plot doesn’t look right at certain viewing angles
This is probably the most commonly reported issue with mplot3d. The problem is that – from some viewing angles – a 3D object would appear in front of another object, even though it is physically behind it. This can result in plots that do not look “physically correct.”
Unfortunately, while some work is being done to reduce the occurrence of this artifact, it is currently an intractable problem, and can not be fully solved until matplotlib supports 3D graphics rendering at its core.
The problem occurs due to the reduction of 3D data down to 2D + z-order scalar. A single value represents the 3rd dimension for all parts of 3D objects in a collection. Therefore, when the bounding boxes of two collections intersect, it becomes possible for this artifact to occur. Furthermore, the intersection of two 3D objects (such as polygons or patches) can not be rendered properly in matplotlib’s 2D rendering engine.
This problem will likely not be solved until OpenGL support is added to all of the backends (patches are greatly welcomed). Until then, if you need complex 3D scenes, we recommend using MayaVi.
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.