Question

I want to use matplotlib to draw more or less the figure I attached below, which includes the two intersecting planes with the right amount of transparency indicating their relative orientations, and the circles and vectors in the two planes projected in 2D.

I’m not sure if there is an existing package for doing this, any hints?

Answer 1

from mpl_toolkits.mplot3d import axes3d
import matplotlib.pyplot as plt
import numpy as np


fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

dim = 10

X, Y = np.meshgrid([-dim, dim], [-dim, dim])
Z = np.zeros((2, 2))

angle = .5
X2, Y2 = np.meshgrid([-dim, dim], [0, dim])
Z2 = Y2 * angle
X3, Y3 = np.meshgrid([-dim, dim], [-dim, 0])
Z3 = Y3 * angle

r = 7
M = 1000
th = np.linspace(0, 2 * np.pi, M)

x, y, z = r * np.cos(th),  r * np.sin(th), angle * r * np.sin(th)

ax.plot_surface(X2, Y3, Z3, color='blue', alpha=.5, linewidth=0, zorder=-1)

ax.plot(x[y < 0], y[y < 0], z[y < 0], lw=5, linestyle='--', color='green',
        zorder=0)

ax.plot_surface(X, Y, Z, color='red', alpha=.5, linewidth=0, zorder=1)

ax.plot(r * np.sin(th), r * np.cos(th), np.zeros(M), lw=5, linestyle='--',
        color='k', zorder=2)

ax.plot_surface(X2, Y2, Z2, color='blue', alpha=.5, linewidth=0, zorder=3)

ax.plot(x[y > 0], y[y > 0], z[y > 0], lw=5, linestyle='--', color='green',
        zorder=4)

plt.axis('off')
plt.show()

caveats:

• I am running a version very close to the current master, so I am not
  sure what will work in older versions

• The reason for splitting up the plotting is that ‘above’ and ‘below’ are determined in a some what arcane way (I am not strictly sure the zorder actually does anything), and is really dependent on the order the artists are drawn in. Thus surfaces can not intersect (one will be above the other every where), so you need to plot the sections on either side of the intersection separately. (You can see this in the black line which I didn’t split at looks like it in ‘on top of’ the upper blue plane).

• The ‘proper’ ordering of the surfaces also seems to be dependent on the view angle.