计算由 scipy 的 Delaunay 函数生成的四面体网格的雅可比矩阵

Question

我正在尝试使用 scipy 的函数 Delaynay 生成四面体网格。根据提供的源代码here，我做了如下的事情：

import math
import random
import numpy as np
from scipy.spatial import Delaunay

# geometry
height = 2.0
depth  = 2.0
width  = 2.0

# random transformations of the original mesh
random.seed(1234)
variation = .00 # as a decimal for a percent

h = 0.4

x_layer = int(width / h)+1
y_layer = int(depth / h)+1
z_layer = int(height / h)+1

points = np.zeros([x_layer*y_layer*z_layer,3])

count = 0
for orig_x in range(x_layer):
    for orig_y in range(y_layer):
        for orig_z in range(z_layer):

            rand_pert_x = random.uniform(-variation, variation)
            rand_pert_y = random.uniform(-variation, variation)
            rand_pert_z = random.uniform(-variation, variation)
    
            x = orig_x*h
            y = orig_y*h
            z = orig_z*h
     
            if (x==0 or orig_x==x_layer-1) or (y==0 or orig_y==y_layer-1) or (z==0 or orig_z==z_layer-1):
                points[count,:] = np.array([x,y,z])
            else:
                points[count,:] = np.array([x+h*rand_pert_x,y+h*rand_pert_y, z+h*rand_pert_z])
            count += 1

tet = Delaunay(points, qhull_options="Qt")

即计划是随机扰动点以创建低质量的网格。现在，variation 设置为 0.0，因此代码将生成 3D 矩形网格。

我想测试网格的质量，所以我正在查看的一件事是每个元素的雅可比行列式。有没有办法从 Delaunay 函数的输出中计算出它？

Answer 1

以下代码计算一个数组

假设一个四面体的点pj=(xj, yj, zj)为j=0, 1, 2, 3，其对应的雅可比矩阵为（见here举例）：

   [[x1-x0, x2-x0, x3-x0], 
J = [y1-y0, y2-y0, y3-y0], 
    [z1-z0, z2-z0, z3-z0]]

下面的函数returns每个四面体对应的雅可比矩阵数组和每个矩阵的行列式值

def compute_delaunay_tetra_jacobians(dt):
    """
    Compute the Jacobian matrix and its determinant for each tetrahedron in the Delaunay triangulation.
    :param dt: the Delaunay triangulation
    :return: array of shape (n, 3, 3) of jacobian matrices such that jacoboian_array[i, :, :] is the 3x3 Jacobian matrix
             array of n values of the determinant of the jacobian matrix
    """
    simp_pts = dt.points[dt.simplices]
    # (n, 4, 3) array of tetrahedra points where simp_pts[i, j, k] holds the k'th coordinate (x, y or z)
    # of the j'th 3D point (of four) of the i'th tetrahedron
    assert simp_pts.shape[1] == 4 and simp_pts.shape[2] == 3

    # building the 3x3 jacobian matrix with entries:
    # [[x1-x0, x2-x0, x3-x0], 
    #  [y1-y0, y2-y0, y3-y0], 
    #  [z1-z0, z2-z0, z3-z0]]
    # 
    a = simp_pts[:, 1, 0] - simp_pts[:, 0, 0]  # x1-x0
    b = simp_pts[:, 1, 1] - simp_pts[:, 0, 1]  # y1-y0
    c = simp_pts[:, 1, 2] - simp_pts[:, 0, 2]  # z1-z0
    d = simp_pts[:, 2, 0] - simp_pts[:, 0, 0]  # x2-x0
    e = simp_pts[:, 2, 1] - simp_pts[:, 0, 1]  # y2-y0
    f = simp_pts[:, 2, 2] - simp_pts[:, 0, 2]  # z2-z0
    g = simp_pts[:, 3, 0] - simp_pts[:, 0, 0]  # x3-x0
    h = simp_pts[:, 3, 1] - simp_pts[:, 0, 1]  # y3-y0
    i = simp_pts[:, 3, 2] - simp_pts[:, 0, 2]  # z3-z0

    determinants = a*(e*i - f*h) - b*(d*i - f*g) + c*(d*h - e*g)

    n = simp_pts.shape[0]
    jacobian_array = np.empty((n, 3, 3))
    jacobian_array[:, 0, 0] = a
    jacobian_array[:, 0, 1] = b
    jacobian_array[:, 0, 2] = c
    jacobian_array[:, 1, 0] = d
    jacobian_array[:, 1, 1] = e
    jacobian_array[:, 1, 2] = f
    jacobian_array[:, 2, 0] = g
    jacobian_array[:, 2, 1] = h
    jacobian_array[:, 2, 2] = i

    return jacobian_array, determinants

计算由 scipy 的 Delaunay 函数生成的四面体网格的雅可比矩阵

Calculate the Jacobian of the tetrahedral mesh generated from scipy's Delaunay function

python

scipy

delaunay