Python：圆柱体（或任何周期表面）的 Delaunay 三角剖分

Question

一年前，我询问了如何在平面上正确地对周期性形状进行三角剖分：环面（对环面进行正确的 Delaunay 三角剖分（使用 python））。

我现在想将其扩展到对圆柱体（或一般来说，任何周期性表面）进行三角测量。我尝试直接扩展二维码：

from scipy.spatial import Delaunay
NZ = 14
NTheta = 14

R = 1 #radius of cylinder 
L = 3 #length of cylinder 

#define base rectangle (u,v)
u=np.linspace(0, 2*np.pi, NTheta) #periodic direction
v=np.linspace(0, L, NZ)
# u=u[:-1] #leave out one point
u,v=np.meshgrid(u,v)
u=u.flatten()
v=v.flatten()

#evaluate the parameterization at the flattened u and v
x=R*np.cos(u)
y=R*np.sin(u)
z=v

#define 2D points, as input data for the Delaunay triangulation of U
points2D=np.vstack([u,v]).T
tri = Delaunay(points2D, incremental=True)#triangulate the rectangle U
triSimplices = tri.simplices

xyz0 = np.vstack([x,y,z]).T

我通过参数化创建一个圆柱体，并通过基本域（矩形）的

scipy.spatial.Delaunay()

获得三角剖分。显然，这个三角测量不知道周期性。通过移动最后一行并绘制，我可以清楚地看到这一点：

为了纠正这个问题，我尝试对 2D 解决方案进行直接扩展 - 在 3D 中添加一个额外的点，重新三角化并删除不需要的单纯形。

Tri1 = Delaunay(points2D) #triangulate the rectangle U
Tri2 = Delaunay(xyz0)

## we add a central (0,0,L/2) point to xy0 to fill it up with triangles
last_pt = xyz0.shape[0]
xy1 = np.vstack((xyz0,(0,0,L/2)))  # add ctr point
Tri3 = Delaunay(xyz1)
print(Tri3.points.shape, Tri3.simplices.shape)
print(Tri1.points.shape, Tri1.simplices.shape)
print(Tri2.points.shape, Tri2.simplices.shape)

## remove the simplices that contain the central point
mask = ~(Tri3.simplices==last_pt).any(axis=1)
triSimplices = Tri3.simplices[mask,:]

然而，将2D代码扩展到3D似乎有一个大问题——3D中的三角剖分给出的是四面体，而不是三角形！而且，似乎对中心点的选择更加敏感。简而言之，我被困住了。

那么，对这样一个周期性表面进行三角测量的正确方法是什么？

Answer 1

以下简短脚本创建顶点，并使用 scipy 创建三角形。然后我重新分配最后一列

# Create a 2d map of the vertices
num_tri_high = 10
num_tri_in_circum = 9
xx = np.linspace(0, 10, num_tri_in_circum + 1)
yy = np.linspace(0, 10, num_tri_high)
xxx,yyy = np.meshgrid(xx,yy)

# Triangulate vertices
receiver_tri = Delaunay(np.transpose(np.vstack((xxx.flatten(), yyy.flatten()))))
triangles = receiver_tri.simplices

# Set last column equal to 1st (which effectively closes the gap)
tot = (num_tri_in_circum + 1) * num_tri_high
for i in range(num_tri_high+1):
    triangles[triangles == (num_tri_in_circum+1) * i + num_tri_in_circum] = (num_tri_in_circum+1) * i

# Wrap into a cylinder
radius = 2
xp = radius*np.sin(xxx.flatten()/np.max(xxx.flatten())*np.pi*2)
yp = radius*np.cos(xxx.flatten()/np.max(xxx.flatten())*np.pi*2)
zp = yyy.flatten()
tow_scene = mlab.triangular_mesh(xp.flatten(),yp.flatten(),zp.flatten(), triangles, representation='wireframe', color=(0,0,0), opacity=1)
mlab.show()