绘制锥体时断开连接的曲面

Question

我想用python绘制曲面(z + 1)²=x²+y²和4z =x²+y².

我写了这段代码:

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')
X= np.arange(-2,3,.1)
Z=np.arange(0,2,.1)
X,Z = np.meshgrid(X,Z)
Y=np.sqrt((Z+1)**2-X**2)
Y2=np.sqrt(4*Z-X**2)
ax.plot_wireframe(X, Y, Z, rstride = 1, cstride =1)
ax.plot_wireframe(X, -Y, Z, rstride = 1, cstride =1)
ax.plot_surface(X,Y2,Z,rstride=1,cstride=1,color='red')
ax.plot_surface(X,-Y2,Z,rstride=1,cstride=1,color='red')
ax.set_zlim(0,2)

plt.show()

这必须显示两个锥体.但是,每个锥体都不是连续的,即有一些面缺失,我不知道为什么.任何帮助将非常感谢.

Answer 1

你定义X和Y的方式在这些连接上引起了一些惊愕.通过在将半径和角度转换为X和Y之前定义锥体,可以获得更平滑的连接,这样您就可以保持以旧方式生成的漂亮Z轮廓.

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')

# Set up the grid in polar
theta = np.linspace(0,2*np.pi,90)
r = np.linspace(0,3,50)
T, R = np.meshgrid(theta, r)

# Then calculate X, Y, and Z
X = R * np.cos(T)
Y = R * np.sin(T)
Z = np.sqrt(X**2 + Y**2) - 1

# Set the Z values outside your range to NaNs so they aren't plotted
Z[Z < 0] = np.nan
Z[Z > 2.1] = np.nan
ax.plot_wireframe(X, Y, Z)

ax.set_zlim(0,2)

plt.show()

这会给你一个非常好的锥形: