如何绘制具有非恒定半径的圆柱体

Question

我编写了代码来生成一个具有恒定固定半径的圆柱体：

import numpy as np
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from math import sin, cos, pi

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

theta = np.linspace(-2*pi,2*pi, 600)
Z = np.linspace(0,1,700)

Z,theta = np.meshgrid(Z, theta)

R = 0.1
X = (R*np.cos(theta))
Y = (R*np.sin(theta))

ax.plot_surface(X,Y,Z,linewidth = 0,facecolor = 'r', shade = True, alpha = 0.6)
plt.show()

如何更改此值以使圆柱半径可以变化。 例如，圆柱体一端的半径从0.1开始，每个连续的“圆”绘制的半径比之前的半径大0.01？ 换句话说，我希望将具有不同半径的圆圈“连接”在一起，形成一个半径不等的圆柱体。

Answer 1

改变了R = np.linspace(0,1,700)

import numpy as np
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from math import sin, cos, pi

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

theta = np.linspace(-2*pi,2*pi, 600)
Z = np.linspace(0,1,700)

Z,theta = np.meshgrid(Z, theta)

R = np.linspace(0,1,700)
X = (R*np.cos(theta))
Y = (R*np.sin(theta))

ax.plot_surface(X,Y,Z,linewidth = 0,facecolor = 'r', shade = True, alpha = 0.6)
plt.show()

你也可以尝试像R = np.sin(np.linspace(0,1,700) * 4) + 1这样的函数