python matplotlib:绘制带圆周的 3D 球体 [英] python matplotlib: drawing 3D sphere with circumferences

问题描述

我正在尝试使用 matplotlib 绘制一个像这样的球体:

I'm trying to draw a sphere like this one using matplotlib:

但是我找不到在背面有虚线的方法，而且垂直圆周看起来有点奇怪

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')

推荐答案

在您显示的示例中，我认为圆不能彼此垂直(即，一个圆是赤道，一个圆角穿过北极)和南极).如果水平圆圈是赤道，则北极必须位于通过代表球体的黄色圆圈中心绘制的垂直线上的某个位置.否则，赤道的右侧将看起来比左侧高或低.但是，表示极圆的椭圆仅穿过黄色圆顶部和底部的中心线.因此，北极位于球体的顶部，这意味着我们必须在赤道上直视，这意味着它应该看起来像一条线，而不是椭圆.

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.

这里有一些代码可以重现与您发布的图类似的内容:

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()

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