ICP-通过一组匹配的3D点估计相机运动

发布于:2025-03-01 ⋅ 阅读:(10) ⋅ 点赞:(0)

问题描述

假设有两组已匹配的3D点:

  • 源点云 { p i ∈ R 3 } i = 1 N \{ \mathbf{p}_i \in \mathbb{R}^3 \}_{i=1}^N {piR3}i=1N
  • 目标点云 { q i ∈ R 3 } i = 1 N \{ \mathbf{q}_i \in \mathbb{R}^3 \}_{i=1}^N {qiR3}i=1N

目标是求解刚体变换旋转矩阵 R ∈ SO ( 3 ) \mathbf{R} \in \text{SO}(3) RSO(3) 和平移向量 t ∈ R 3 \mathbf{t} \in \mathbb{R}^3 tR3,使得变换后的源点云与目标点云对齐,即最小化以下目标函数:
min ⁡ R , t ∑ i = 1 N ∥ R p i + t − q i ∥ 2 . \min_{\mathbf{R}, \mathbf{t}} \sum_{i=1}^N \left\| \mathbf{R} \mathbf{p}_i + \mathbf{t} - \mathbf{q}_i \right\|^2. R,tmini=1NRpi+tqi2.

SVD方法

分析计算

1. 去中心化与平移向量的求解

首先计算两组点的质心:
μ p = 1 N ∑ i = 1 N p i , μ q = 1 N ∑ i = 1 N q i . \boldsymbol{\mu}_p = \frac{1}{N} \sum_{i=1}^N \mathbf{p}_i, \quad \boldsymbol{\mu}_q = \frac{1}{N} \sum_{i=1}^N \mathbf{q}_i. μp=N1i=1Npi,μq=N1i=1Nqi.
将每个点减去质心,得到去中心化的坐标:
x i = p i − μ p , y i = q i − μ q . \mathbf{x}_i = \mathbf{p}_i - \boldsymbol{\mu}_p, \quad \mathbf{y}_i = \mathbf{q}_i - \boldsymbol{\mu}_q. xi=piμp,yi=qiμq.
此时,目标函数可简化为:
min ⁡ R , t ∑ i = 1 N ∥ R x i − y i + ( t − μ q + R μ p ) ∥ 2 . \min_{\mathbf{R}, \mathbf{t}} \sum_{i=1}^N \left\| \mathbf{R} \mathbf{x}_i - \mathbf{y}_i + (\mathbf{t} - \boldsymbol{\mu}_q + \mathbf{R} \boldsymbol{\mu}_p) \right\|^2. R,tmini=1N Rxiyi+(tμq+Rμp) 2.
令括号内项为零,可得平移向量的最优解:
t = μ q − R μ p . \mathbf{t} = \boldsymbol{\mu}_q - \mathbf{R} \boldsymbol{\mu}_p. t=μqRμp.
代入后,优化问题简化为仅关于旋转矩阵 (\mathbf{R}) 的最小化问题:
min ⁡ R ∑ i = 1 N ∥ R x i − y i ∥ 2 . \min_{\mathbf{R}} \sum_{i=1}^N \left\| \mathbf{R} \mathbf{x}_i - \mathbf{y}_i \right\|^2. Rmini=1NRxiyi2.


2. 目标函数的展开与迹形式转换

展开平方误差:
∑ i = 1 N ∥ R x i − y i ∥ 2 = ∑ i = 1 N ( R x i − y i ) ⊤ ( R x i − y i ) . \sum_{i=1}^N \left\| \mathbf{R} \mathbf{x}_i - \mathbf{y}_i \right\|^2 = \sum_{i=1}^N \left( \mathbf{R} \mathbf{x}_i - \mathbf{y}_i \right)^\top \left( \mathbf{R} \mathbf{x}_i - \mathbf{y}_i \right). i=1NRxiyi2=i=1N(Rxiyi)(Rxiyi).
进一步展开并利用迹的线性性质:
= ∑ i = 1 N ( x i ⊤ R ⊤ R x i − 2 y i ⊤ R x i + y i ⊤ y i ) . = \sum_{i=1}^N \left( \mathbf{x}_i^\top \mathbf{R}^\top \mathbf{R} \mathbf{x}_i - 2 \mathbf{y}_i^\top \mathbf{R} \mathbf{x}_i + \mathbf{y}_i^\top \mathbf{y}_i \right). =i=1N(xiRRxi2yiRxi+yiyi).
由于 R ⊤ R = I \mathbf{R}^\top \mathbf{R} = \mathbf{I} RR=I,第一项简化为 x i ⊤ x i \mathbf{x}_i^\top \mathbf{x}_i xixi,与 R \mathbf{R} R 无关。优化问题等价于最大化交叉项:
max ⁡ R ∑ i = 1 N y i ⊤ R x i = max ⁡ R tr ( R ⊤ ∑ i = 1 N y i x i ⊤ ) . \max_{\mathbf{R}} \sum_{i=1}^N \mathbf{y}_i^\top \mathbf{R} \mathbf{x}_i = \max_{\mathbf{R}} \text{tr}\left( \mathbf{R}^\top \sum_{i=1}^N \mathbf{y}_i \mathbf{x}_i^\top \right). Rmaxi=1NyiRxi=Rmaxtr(Ri=1Nyixi).
定义协方差矩阵:
H = ∑ i = 1 N y i x i ⊤ . \mathbf{H} = \sum_{i=1}^N \mathbf{y}_i \mathbf{x}_i^\top. H=i=1Nyixi.


3. 通过SVD求解旋转矩阵

对协方差矩阵 H \mathbf{H} H 进行奇异值分解(SVD):
H = U Σ V ⊤ . \mathbf{H} = \mathbf{U} \mathbf{\Sigma} \mathbf{V}^\top. H=V.
目标函数可写为:
tr ( R ⊤ H ) = tr ( R ⊤ U Σ V ⊤ ) . \text{tr}(\mathbf{R}^\top \mathbf{H}) = \text{tr}(\mathbf{R}^\top \mathbf{U} \mathbf{\Sigma} \mathbf{V}^\top). tr(RH)=tr(RV).
利用迹的循环置换性质 ( tr ( A B C ) = tr ( B C A ) (\text{tr}(ABC) = \text{tr}(BCA) (tr(ABC)=tr(BCA),可重写为:
tr ( V ⊤ R ⊤ U Σ ) . \text{tr}(\mathbf{V}^\top \mathbf{R}^\top \mathbf{U} \mathbf{\Sigma}). tr(VR).
M = V ⊤ R ⊤ U \mathbf{M} = \mathbf{V}^\top \mathbf{R}^\top \mathbf{U} M=VRU,则目标函数简化为:
tr ( M Σ ) . \text{tr}(\mathbf{M} \mathbf{\Sigma}). tr().
由于 M \mathbf{M} M 是正交矩阵 ( M ⊤ M = I (\mathbf{M}^\top \mathbf{M} = \mathbf{I} (MM=I),其对角线元素绝对值不超过1。为了使 tr ( M Σ ) \text{tr}(\mathbf{M} \mathbf{\Sigma}) tr() 最大,需使 M \mathbf{M} M的对角线元素尽可能大。 M = I \mathbf{M} = \mathbf{I} M=I,迹达到最大值 tr ( Σ ) \text{tr}(\mathbf{\Sigma}) tr(Σ),即:
M = I    ⟹    V ⊤ R ⊤ U = I . \mathbf{M} = \mathbf{I} \implies \mathbf{V}^\top \mathbf{R}^\top \mathbf{U} = \mathbf{I}. M=IVRU=I.
解得旋转矩阵:
R ⊤ = V U ⊤    ⟹    R = U V ⊤ . \mathbf{R}^\top = \mathbf{V} \mathbf{U}^\top \implies \mathbf{R} = \mathbf{U} \mathbf{V}^\top. R=VUR=UV.

验证行列式

  • det ⁡ ( R ) = 1 \det(\mathbf{R}) = 1 det(R)=1,解有效。
  • det ⁡ ( R ) = − 1 \det(\mathbf{R}) = -1 det(R)=1(存在反射),需修正为:
    R = V [ 1 0 0 0 1 0 0 0 − 1 ] U ⊤ . \mathbf{R} = \mathbf{V} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \mathbf{U}^\top. R=V 100010001 U.

4. 公式总结
  1. 质心计算
    μ p = 1 N ∑ i = 1 N p i , μ q = 1 N ∑ i = 1 N q i . \boldsymbol{\mu}_p = \frac{1}{N} \sum_{i=1}^N \mathbf{p}_i, \quad \boldsymbol{\mu}_q = \frac{1}{N} \sum_{i=1}^N \mathbf{q}_i. μp=N1i=1Npi,μq=N1i=1Nqi.
  2. 协方差矩阵
    H = ∑ i = 1 N ( q i − μ q ) ( p i − μ p ) ⊤ . \mathbf{H} = \sum_{i=1}^N (\mathbf{q}_i - \boldsymbol{\mu}_q)(\mathbf{p}_i - \boldsymbol{\mu}_p)^\top. H=i=1N(qiμq)(piμp).
  3. SVD分解与旋转矩阵
    H = U Σ V ⊤ , R = V U ⊤ ( 确保 det ⁡ ( R ) = 1 ) . \mathbf{H} = \mathbf{U} \mathbf{\Sigma} \mathbf{V}^\top, \quad \mathbf{R} = \mathbf{V} \mathbf{U}^\top \quad (\text{确保} \det(\mathbf{R}) = 1). H=V,R=VU(确保det(R)=1).
  4. 平移向量
    t = μ q − R μ p . \mathbf{t} = \boldsymbol{\mu}_q - \mathbf{R} \boldsymbol{\mu}_p. t=μqRμp.

几何解释

通过SVD分解,协方差矩阵 H \mathbf{H} H 的奇异向量反映了点云之间的主要对齐方向。最优旋转矩阵 R \mathbf{R} R 通过最大化点对之间的协方差,使得变换后的源点云与目标点云在最小二乘意义下对齐。平移向量 t \mathbf{t} t 则通过质心差校正整体偏移。


算法步骤总结

  1. 去中心化:计算质心并减去,得到 x i \mathbf{x}_i xi y i \mathbf{y}_i yi
  2. 构建协方差矩阵 H = ∑ y i x i ⊤ \mathbf{H} = \sum \mathbf{y}_i \mathbf{x}_i^\top H=yixi
  3. SVD分解 H = U Σ V ⊤ \mathbf{H} = \mathbf{U} \mathbf{\Sigma} \mathbf{V}^\top H=V
  4. 求解旋转矩阵 R = V U ⊤ \mathbf{R} = \mathbf{V} \mathbf{U}^\top R=VU,调整行列式。
  5. 计算平移向量 t = μ q − R μ p \mathbf{t} = \boldsymbol{\mu}_q - \mathbf{R} \boldsymbol{\mu}_p t=μqRμp

代码示例(Python)

import numpy as np

def solve_icp(source_points, target_points):
    # 输入:source_points (n×3), target_points (n×3)
    # 输出:R (3×3), t (3×1)
    
    # 计算质心
    mu_p = np.mean(source_points, axis=0)
    mu_q = np.mean(target_points, axis=0)
    
    # 去中心化
    X = source_points - mu_p
    Y = target_points - mu_q
    
    # 协方差矩阵
    H = X.T @ Y
    
    # SVD分解
    U, S, Vt = np.linalg.svd(H)
    
    # 计算旋转矩阵
    R = Vt.T @ U.T
    
    # 处理反射情况
    if np.linalg.det(R) < 0:
        Vt[-1, :] *= -1
        R = Vt.T @ U.T
    
    # 计算平移向量
    t = mu_q - R @ mu_p
    
    return R, t

# 示例数据
source = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
target = np.array([[2, 3, 4], [5, 6, 7], [8, 9, 10]])

R, t = solve_icp(source, target)
print("Rotation Matrix:\n", R)
print("Translation Vector:\n", t)

关键点

  1. 闭式解:通过SVD直接求解最优变换,无需迭代(适用于精确匹配)。
  2. 反射问题:需检查 det ⁡ ( R ) \det(\mathbf{R}) det(R),避免反射变换。
  3. 迭代优化:若存在噪声或匹配误差,需多次迭代优化。

通过上述步骤,可高效求解已知匹配3D点对的刚体变换,适用于点云配准、物体位姿估计等场景。