我想使用与在Matplotlib中绘制以平面为中心的实心圆柱体相同的方法来绘制截锥体;该方法在已知底部中心上的两个点和半径时绘制圆柱体。另一方面,当已知其底部中心上的两个点的坐标和每个底部的半径时,我想绘制一个截锥体。看起来我只需要更改以下程序中绘制圆柱体的倒数第二行,但是我尝试了很多次都无法做到。
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
from scipy.linalg import norm
import pylab as pllt
fig = pllt.figure()
ax = fig.add_subplot(1,1,1, projection='3d')
#ax = pllt.subplot2grid((2,2), (0,0), rowspan=2, projection='3d')
#axis and radius
def cylinder(p0,p1,R,ccc):
#vector in direction of axis
v = p1 - p0
#find magnitude of vector
mag = norm(v)
#unit vector in direction of axis
v = v / mag
#make some vector not in the same direction as v
not_v = np.array([1, 1, 0])
if (v == not_v).all():
not_v = np.array([0, 1, 0])
#make vector perpendicular to v
n1 = np.cross(v, not_v)
#print n1,'\t',norm(n1)
#normalize n1
n1 /= norm(n1)
#make unit vector perpendicular to v and n1
n2 = np.cross(v, n1)
#surface ranges over t from 0 to length of axis and 0 to 2*pi
t = np.linspace(0, mag, 80)
theta = np.linspace(0, 2 * np.pi, 80)
#use meshgrid to make 2d arrays
t, theta = np.meshgrid(t, theta)
#generate coordinates for surface
X, Y, Z = [p0[i] + v[i] * t + R * np.sin(theta) * n1[i] + R * np.cos(theta) * n2[i] for i in [0, 1, 2]]
ax.plot_surface(X, Y, Z,color=ccc,linewidth=0, antialiased=False)
A0 = np.array([1, 3, 2])
A1 = np.array([8, 5, 9])
ax.set_xlim(0,10)
ax.set_ylim(0,10)
ax.set_zlim(0,10)
cylinder(A0,A1,1,'blue')
pllt.show()
我认为我应该按照http://mathworld.wolfram.com/ConicalFrustum.html中提到的v=p1-p0的函数来改变半径,以便能够完成此操作。 如果有任何方法,请告诉我。