Question

我有一个定义的平面（恰好与3D空间中两个xyz点定义的向量正交）。我可以将任何xyz点投影到平面上并表示该投影uv坐标空间。我想在uv坐标空间中取任意点，并找出其在xyz空间中的坐标。

a = x2 - x1
b = y2 - y1
c = z2 - z1
d = -1*(a*x1 + b*y1 + c*z1)
magnitude = (a**2 + b**2 + c**2)**.5
u_magnitude = (b**2 + a**2)**.5

normal = [a/magnitude, b/magnitude, c/magnitude]
u = [b/u_magnitude, -a/u_magnitude, 0]
v = np.cross(normal, u)

p_u = np.dot(u,[x1,y1,z1])
p_v = np.dot(v,[x1,y1,z1])

我相信此代码可以准确地生成所需的平面，并将uv坐标中的x1，y1，z1点分配给p_u，p_v。我的感觉是，我拥有执行反向操作所需的一切，但我不知道如何做。如果我有一个点u0，v0，如何找到描述其在3D空间中位置的x0，y0，z0？

Answer 1

根据文本中的定义（不阅读代码），问题没有得到很好的定义-因为与给定向量正交的平面数量无限（将所有选项视为沿不同“偏移”的平面）从第一点到第二点的线）。您需要首先选择飞机必须经过的点。

第二，当我们将（U，V）对转换为3D点时，我假设您是指飞机上的3D点。

不过，尝试更具体一点，这是您的代码，以及有关我如何理解它以及如何进行相反操作的文档：

# ### The original computation of the plane equation ###
# Given points p1 and p2, the vector through them is W = (p2 - p1)
# We want the plane equation Ax + By + Cz + d = 0, and to make
# the plane prepandicular to the vector, we set (A, B, C) = W
p1 = np.array([x1, y1, z1])
p2 = np.array([x2, y2, z2])

A, B, C = W = p2 - p1

# Now we can solve D in the plane equation. This solution assumes that
# the plane goes through p1.
D = -1 * np.dot(W, p1)

# ### Normalizing W ###
magnitude = np.linalg.norm(W)
normal = W / magnitude

# Now that we have the plane, we want to define
# three things:
# 1. The reference point in the plane (the "origin"). Given the
#    above computation of D, that is p1.
# 2. The vectors U and V that are prepandicular to W
#    (and therefore spanning the plane)

# We take a vector U that we know that is perpendicular to
# W, but we also need to make sure it's not zero.
if A != 0:
    u_not_normalized = np.array([B, -A, 0])
else:
    # If A is 0, then either B or C have to be nonzero
    u_not_normalized = np.array([0, B, -C])
u_magnitude = np.linalg.norm(u_not_normalized)

# ### Normalizing W ###
U = u_not_normalized / u_magnitude
V = np.cross(normal, U)

# Now, for a point p3 = (x3, y3, z3) it's (u, v) coordinates would be
# computed relative to our reference point (p1)
p3 = np.array([x3, y3, z3])

p3_u = np.dot(U, p3 - p1)
p3_v = np.dot(V, p3 - p1)

# And to convert the point back to 3D, we just use the same reference point
# and multiply U and V by the coordinates
p3_again = p1 + p3_u * U + p3_v * V

如何将平面uv坐标中定义平面上的2D点转换回3D xyz坐标？

1 个答案: