I have managed to create a code that outputs a sphere and a ring around it but the ring doesn't look like its rounding the sphere...
Can anyone save me here?
Code:
import numpy as np
from matplotlib import pyplot as plt
import mpl_toolkits.mplot3d.axes3d
theta = np.linspace(0, 2.*np.pi, 100)
phi = np.linspace(0, 2*np.pi, 100)
theta, phi = np.meshgrid(theta, phi)
c, a = 10, 0.2
x = (c + a*np.cos(theta)) * np.cos(phi)
y = (c + a*np.cos(theta)) * np.sin(phi)
z = a * np.sin(theta)
fig = plt.figure()
ax1 = fig.add_subplot(111, projection='3d')
ax1.set_zlim(-10,10)
ax1.plot_surface(x, y, z, rstride=5, cstride=5, color='b', edgecolors='b')
ax1.view_init(36, 26)
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)
x2 = 4 * np.outer(np.cos(u), np.sin(v))
y2 = 4 * np.outer(np.sin(u), np.sin(v))
z2 = 10 * np.outer(np.ones(np.size(u)), np.cos(v))
ax1.plot_surface(x2, y2, z2, rstride=5, cstride=5, color='r')
plt.show()
Related
I have Python code to plot solid of revolution from function. How can I modify this to be able to create solid of revolution (revolved at certain x or y axis that we can choose. Precisely for these:
2n-regular polygons inscribed in a circle
Triangle, not only right angle, any kind of triangle
If it is a circle then it will become a torus
# Compare the plot at xy axis with the solid of revolution toward x and y axis
# For region bounded by the line x - 2y = 0 and y^2 = 4x
# Plotting the revolution of the bounded region
# can be done by limiting the np.linspace of the y, u, and x_inverse values
# You can determine the limits by finding the intersection points of the two functions.
import matplotlib.pyplot as plt
import numpy as np
import sympy as sy
def r1(x):
return x/2
def r2(x):
return 2*(x**(1/2))
def r3(x):
return 2*x
def r4(x):
return (x/2)**(2)
def vx(x):
return np.pi*(r2(x)**2 - r1(x)**2)
def vy(x):
return np.pi*(r3(x)**2 - r4(x)**2)
x = sy.Symbol("x")
vx = sy.integrate(vx(x), (x, 0, 16))
vy = sy.integrate(vy(x), (x, 0, 8))
n = 200
fig = plt.figure(figsize=(14, 7))
ax1 = fig.add_subplot(221)
ax2 = fig.add_subplot(222, projection='3d')
ax3 = fig.add_subplot(223)
ax4 = fig.add_subplot(224, projection='3d')
y = np.linspace(0, 8, n)
x1 = (2*y)
x2 = (y / 2) ** (2)
t = np.linspace(0, np.pi * 2, n)
u = np.linspace(0, 16, n)
v = np.linspace(0, 2 * np.pi, n)
U, V = np.meshgrid(u, v)
X = U
Y1 = (2 * U ** (1/2)) * np.cos(V)
Z1 = (2 * U ** (1/2)) * np.sin(V)
Y2 = (U / 2) * np.cos(V)
Z2 = (U / 2) * np.sin(V)
Y3 = ((U / 2) ** (2)) * np.cos(V)
Z3 = ((U / 2) ** (2)) * np.sin(V)
Y4 = (2*U) * np.cos(V)
Z4 = (2*U) * np.sin(V)
ax1.plot(x1, y, label='$y=x/2$')
ax1.plot(x2, y, label='$y=2 \sqrt{x}$')
ax1.legend()
ax1.set_title('$f(x)$')
ax2.plot_surface(X, Y3, Z3, alpha=0.3, color='red', rstride=6, cstride=12)
ax2.plot_surface(X, Y4, Z4, alpha=0.3, color='blue', rstride=6, cstride=12)
ax2.set_title("$f(x)$: Revolution around $y$ \n Volume = {}".format(vy))
# find the inverse of the function
x_inverse = np.linspace(0, 8, n)
y1_inverse = np.power(2*x_inverse, 1)
y2_inverse = np.power(x_inverse / 2, 2)
ax3.plot(x_inverse, y1_inverse, label='Inverse of $y=x/2$')
ax3.plot(x_inverse, y2_inverse, label='Inverse of $y=2 \sqrt{x}$')
ax3.set_title('Inverse of $f(x)$')
ax3.legend()
ax4.plot_surface(X, Y1, Z1, alpha=0.3, color='red', rstride=6, cstride=12)
ax4.plot_surface(X, Y2, Z2, alpha=0.3, color='blue', rstride=6, cstride=12)
ax4.set_title("$f(x)$: Revolution around $x$ \n Volume = {}".format(vx))
plt.tight_layout()
plt.show()
I have two formulas:
x^2 + y^2+z^2 = 6
(x-3)^2+y^2=4
So I have to draw this shapes by matplotlib on python.
My solution:
`
import math
import matplotlib.pyplot as plt
from pylab import *
import numpy as np
import matplotlib.patches as patches
from mpl_toolkits.mplot3d import Axes3D
from itertools import product, combinations
import mpl_toolkits.mplot3d.art3d as art3d
def task3():
fig = plt.figure()
ax = plt.axes(projection='3d')
coefs = (1, 1, 1) # Coefficients in a0/c x**2 + a1/c y**2 + a2/c z**2 = 1
# Radii corresponding to the coefficients:
rx, ry, rz = 1 / np.sqrt(coefs)
# Set of all spherical angles:
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)
# Cartesian coordinates that correspond to the spherical angles:
# (this is the equation of an ellipsoid):
x = rx * np.outer(np.cos(u), np.sin(v))
y = ry * np.outer(np.sin(u), np.sin(v))
z = rz * np.outer(np.ones_like(u), np.cos(v))
ax.plot_surface(x, y, z, rstride=6, cstride=6,
cmap='viridis', edgecolor='none')
# ax.scatter(X, Y, Z, c=Z, cmap='viridis', label="xy")
radius = 0.3
height = 2.4
elevation = -1
resolution = 720
color = 'r'
x_center = 0.7
y_center = -0
a = np.linspace(x_center - radius, x_center + radius, resolution)
b = np.linspace(elevation, elevation + height, resolution)
X, Z = np.meshgrid(a, b)
Y = np.sqrt(radius ** 2 - (X - x_center) ** 2) + y_center # Pythagorean theorem
ax.plot_surface(X, Y, Z, rstride=6, cstride=6,
cmap='viridis', edgecolor='none')
ax.plot_surface(X, (2 * y_center - Y), Z, rstride=6, cstride=6,
cmap='viridis', edgecolor='none')
plt.show()
[![Result. I cant post images, but you can compile result and then you will see that my solution too approximate)
I want to do it correctly
I'm trying to draw a sphere like this one using matplotlib:
but I can't find a way of having a dashed lines on the back and the vertical circumference looks a bit strange
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure(figsize=(12,12), dpi=300)
ax = fig.add_subplot(111, projection='3d')
ax.set_aspect('equal')
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)
x = 1 * np.outer(np.cos(u), np.sin(v))
y = 1 * np.outer(np.sin(u), np.sin(v))
z = 1 * np.outer(np.ones(np.size(u)), np.cos(v))
#for i in range(2):
# ax.plot_surface(x+random.randint(-5,5), y+random.randint(-5,5), z+random.randint(-5,5), rstride=4, cstride=4, color='b', linewidth=0, alpha=0.5)
ax.plot_surface(x, y, z, rstride=4, cstride=4, color='b', linewidth=0, alpha=0.5)
ax.plot(np.sin(theta),np.cos(u),0,color='k')
ax.plot([0]*100,np.sin(theta),np.cos(u),color='k')
In the example you show, I don't think that the circles can be perpendicular to one another (i.e. one is the equator and one runs through the north pole and south pole). If the horizontal circle is the equator, then the north pole must be somewhere on a vertical line drawn through the center of the yellow circle that represents the sphere. Otherwise, the right side of the equator would look higher or lower than the left. However, the ellipse that represents the polar circle only crosses through that center line at the top and bottom of the yellow circle. Therefore, the north pole is at the top of the sphere, which means that we must be looking straight on the equator, which means it should look like a line, not an ellipse.
Here's some code to reproduce something similar to the figure you posted:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.set_aspect('equal')
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)
x = 1 * np.outer(np.cos(u), np.sin(v))
y = 1 * np.outer(np.sin(u), np.sin(v))
z = 1 * np.outer(np.ones(np.size(u)), np.cos(v))
#for i in range(2):
# ax.plot_surface(x+random.randint(-5,5), y+random.randint(-5,5), z+random.randint(-5,5), rstride=4, cstride=4, color='b', linewidth=0, alpha=0.5)
elev = 10.0
rot = 80.0 / 180 * np.pi
ax.plot_surface(x, y, z, rstride=4, cstride=4, color='b', linewidth=0, alpha=0.5)
#calculate vectors for "vertical" circle
a = np.array([-np.sin(elev / 180 * np.pi), 0, np.cos(elev / 180 * np.pi)])
b = np.array([0, 1, 0])
b = b * np.cos(rot) + np.cross(a, b) * np.sin(rot) + a * np.dot(a, b) * (1 - np.cos(rot))
ax.plot(np.sin(u),np.cos(u),0,color='k', linestyle = 'dashed')
horiz_front = np.linspace(0, np.pi, 100)
ax.plot(np.sin(horiz_front),np.cos(horiz_front),0,color='k')
vert_front = np.linspace(np.pi / 2, 3 * np.pi / 2, 100)
ax.plot(a[0] * np.sin(u) + b[0] * np.cos(u), b[1] * np.cos(u), a[2] * np.sin(u) + b[2] * np.cos(u),color='k', linestyle = 'dashed')
ax.plot(a[0] * np.sin(vert_front) + b[0] * np.cos(vert_front), b[1] * np.cos(vert_front), a[2] * np.sin(vert_front) + b[2] * np.cos(vert_front),color='k')
ax.view_init(elev = elev, azim = 0)
plt.show()
I am trying to plot an ellipsoid so, I thought I would amend the example code for a sphere from the matplotlib 3D plotting page.
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from matplotlib import cm
import numpy as np
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
# Ellipsoid
u = np.linspace(-np.pi/2.0,np.pi/2.0,100)
v = np.linspace(-np.pi,np.pi,100)
x = 10 * np.outer(np.cos(u), np.cos(v))
y = 10 * np.outer(np.cos(u), np.sin(v))
z = 10 * np.outer(np.ones(np.size(u)), np.sin(v))
# Sphere
#u = np.linspace(0, 2 * np.pi, 100)
#v = np.linspace(0, np.pi, 100)
#x = 10 * np.outer(np.cos(u), np.sin(v))
#y = 10 * np.outer(np.sin(u), np.sin(v))
#z = 10 * np.outer(np.ones(np.size(u)), np.cos(v))
ax.plot_surface(x, y, z, rstride=4, cstride=4, cmap = cm.copper)
ax.set_xlabel('x-axis')
ax.set_ylabel('y-axis')
ax.set_zlabel('z-axis')
plt.show()
If you run the code you will see that the plot returns an aesthetically pleasing half inside out boat like surface but sadly not an ellipsoid.
Have included the sphere code (commented out) for comparison.
Is there something obvious here that I'm missing?
Why did you change the parametrization? Starting with the sphere as an example, you only have to change the semi-axis lengths:
# Ellipsoid
u = np.linspace(0, 2.*np.pi, 100)
v = np.linspace(0, np.pi, 100)
x = 60 * np.outer(np.cos(u), np.sin(v))
y = 20 * np.outer(np.sin(u), np.sin(v))
z = 10 * np.outer(np.ones(np.size(u)), np.cos(v))
I would like to graph a 3d plot of a sphere, which colors sections differently according to some function of the theta, phi coordinates. I could graph two separate graphs, one of each color, but I'm not sure exactly how the gridding/meshing works when plotting. Viz. I want to grid/mesh over all thetas/phis in a sphere, but then throw out certain pairs given a boolean function. Is this possible? Attached is a picture of a scatterplot that does what I'd like to do with surfaces.
Based on scatter3d_demo.py found in the matplotlib tutorial:
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1, projection='3d')
THETA, PHI = np.ogrid[0:2*np.pi:40j, 0:np.pi:30j]
X = 10 * np.cos(THETA) * np.sin(PHI)
Y = 10 * np.sin(THETA) * np.sin(PHI)
Z = 10 * np.ones_like(THETA) * np.cos(PHI)
def func(THETA, PHI):
mask = (THETA < np.pi/2) & (np.pi/3 < PHI) & (PHI < 2 * np.pi/3)
return np.where(mask, 1, 0.5)
C = func(THETA, PHI)
x = X.ravel()
y = Y.ravel()
z = Z.ravel()
c = C.ravel()
ax.scatter(x, y, z, c=c, s=30, vmin=0, vmax=1)
ax.set_aspect('equal')
plt.show()
yields
Note you could also color patches on a sphere using plot_surface:
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
THETA, PHI = np.ogrid[0:2*np.pi:40j, 0:np.pi:30j]
X = 10 * np.cos(THETA) * np.sin(PHI)
Y = 10 * np.sin(THETA) * np.sin(PHI)
Z = 10 * np.ones_like(THETA) * np.cos(PHI)
def func(THETA, PHI):
mask = (THETA < np.pi/2) & (np.pi/3 < PHI) & (PHI < 2 * np.pi/3)
return np.where(mask, 1, 0.5)
C = func(THETA, PHI)
jet = plt.cm.jet
ax.plot_surface(X, Y, Z, rstride=2, cstride=2, facecolors=jet(C))
ax.set_aspect('equal')
plt.show()