对极约束
[oldsymbol{x}_{2}^{T} oldsymbol{F} oldsymbol{x}_{1}=oldsymbol{0} quad hat{oldsymbol{x}}_{2}^{T} oldsymbol{E} hat{oldsymbol{x}}_{1}=mathbf{0}
]
其中
[oldsymbol{E}=oldsymbol{K}_{2}^{-T} oldsymbol{F K}_{1} quad hat{oldsymbol{x}}_{1}=oldsymbol{K}_{1}^{-1} oldsymbol{x}_{1} quad hat{oldsymbol{x}}_{2}=oldsymbol{K}_{2}^{-1} oldsymbol{x}_{2}
]
公式推导
[oldsymbol{P}_{1}=oldsymbol{K}_{1}[oldsymbol{I}, quad mathbf{0}] quad oldsymbol{P}_{2}=oldsymbol{K}_{2} left[ egin{array}{ll}{oldsymbol{R},} & {oldsymbol{t}}end{array}
ight]
]
[d_{1} x_{1}=K_{1} X quad Rightarrow quad d_{1} K_{1}^{-1} x_{1}=X=d_{1} hat{x}_{1}
]
[d_{2} oldsymbol{x}_{2}=oldsymbol{K}_{2}(oldsymbol{R} oldsymbol{X}+oldsymbol{t}) Rightarrow quad d_{2} oldsymbol{K}_{2}^{-1} oldsymbol{x}_{2}=oldsymbol{R} oldsymbol{X}+oldsymbol{t}=d_{1} oldsymbol{R} hat{x}_{1}+oldsymbol{t}
]
[quad d_{2}[t]_{ imes} hat{x}_{2}=d_{1}[t]_{ imes} R hat{x}_{1}+[t]_{ imes} t
]
[quad d_{2} hat{x}_{2}^{T}[t]_{ imes} hat{x}_{2}=d_{1} hat{x}_{2}^{T}[t]_{ imes} R hat{x}_{1}=0
]
[hat{oldsymbol{x}}_{2}^{T}[oldsymbol{t}]_{ imes} oldsymbol{R} hat{x}_{1}=hat{oldsymbol{x}}_{2}^{T} oldsymbol{E} hat{x}_{1}=0
]
[oldsymbol{E}=[oldsymbol{t}]_{ imes} oldsymbol{R}
]
[oldsymbol{x}_{2}^{T} oldsymbol{K}_{2}^{-T}[oldsymbol{t}]_{ imes} oldsymbol{R} oldsymbol{K}_{1}^{-1} oldsymbol{x}_{1}=oldsymbol{x}_{2}^{T} oldsymbol{F} oldsymbol{x}_{1}=0
]
[oldsymbol{F}=oldsymbol{K}_{2}^{-T} oldsymbol{E} oldsymbol{K}_{1}^{-1}
]
基础矩阵性质
-
3x3的矩阵，秩为2
-
具有7个自由度
-
奇异值为(left[ egin{array}{lll}{sigma_{1},} & {sigma_{2},} & {0}end{array} ight]^{T})
-
极线约束(l_{1}=oldsymbol{x}_{2}^{T} oldsymbol{F}, quad l_{2}=oldsymbol{F} oldsymbol{x}_{1}, quad oldsymbol{x}_{2}^{T} oldsymbol{F} oldsymbol{x}_{1}=0)
基础矩阵求解方法
直接线性变换法 （8点法+最小二乘法）
对于一对匹配点
[oldsymbol{x}_{1}=left[ egin{array}{ll}{u_{1},} & {v_{1},} & {1}end{array}
ight]^{mathrm{T}}, quad oldsymbol{x}_{2}=left[ egin{array}{ll}{u_{2},} & {v_{2},} & {1}end{array}
ight]^{mathrm{T}}
]
根据对极约束
[oldsymbol{x}_{2}^{T} oldsymbol{F} oldsymbol{x}_{1}=oldsymbol{0}
]
有
[left( egin{array}{lll}{u_{1}} & {v_{1}} & {1}end{array}
ight) left[ egin{array}{ccc}{F_{11}} & {F_{12}} & {F_{13}} \ {F_{21}} & {F_{22}} & {F_{23}} \ {F_{31}} & {F_{32}} & {F_{33}}end{array}
ight] left( egin{array}{c}{u_{2}} \ {v_{2}} \ {1}end{array}
ight)=0
]
令
[oldsymbol{f}=left[ egin{array}{lll}{F_{11},} & {F_{12},} & {F_{13}}, quad end{array}
ight.F_{21}, quad F_{22}, quad F_{23}, quad F_{31}, quad F_{32}, quad F_{33} ]^{T}
]
有约束
[left[ egin{array}{llllll}{u_{1} u_{1},} & {u_{1} v_{2},} & {u_{1},} & {v_{2} u_{1},} & {v_{1} v_{2},} & {v_{1},} & {u_{2},} & {v_{2},} & {1}end{array}
ight] f=0
]
当有n对匹配点时
[A=left( egin{array}{ccccccccc}
{egin{array}{l}
{u_{1}^{(1)} u_{1}^{(1)}, quad u_{1}^{(1)} v_{2}^{(1)}, quad u_{1}^{(1)}, quad v_{1}^{(1)} u_{2}^{(1)}, quad v_{1}^{(1)} v_{2}^{(1)}, quad v_{1}^{(1)}, quad u_{2}^{(1)}, quad v_{2}^{(1)}, quad 1}
\ {u_{1}^{(2)} u_{1}^{(2)}, quad u_{1}^{(2)} v_{2}^{(2)}, quad u_{1}^{(2)}, quad v_{1}^{(2)} u_{2}^{(2)}, quad v_{1}^{(2)} v_{2}^{(2)}, quad v_{1}^{(2)}, quad u_{2}^{(2)}, quad v_{2}^{(2)}, quad 1}
\ quad vdots quadquad quad vdots quadquad quad vdotsquad quadquad vdots quad quadquad vdots quadquad vdotsquad quad vdotsquadquad quad vdotsquadquad vdots
\u_{1}^{(n)} u_{1}^{(n)}, quad u_{1}^{(n)} v_{2}^{(n)}, quad u_{1}^{(n)}, quad v_{1}^{(n)} u_{2}^{(n)}, quad v_{1}^{(n)} v_{2}^{(n)}, quad v_{1}^{(n)}, quad u_{2}^{(n)}, quad v_{2}^{(n)}, quad 1
end{array}}
end{array}
ight).
]
[A f=0
]
- 要保证有唯一解至少需要8对匹配点
- n = 8时，若A非奇异，则有唯一解，称为8点法
- n >= 8时，可用最小二乘法求解( $ A^TA $ 的最小特征值对应的特征向量即为最优解)
奇异值约束
直接线性变化法无法保证基础矩阵的奇异值约束—有两个非0奇异值
根据奇异值约束对矩阵进行重构
[min |oldsymbol{F}-hat{oldsymbol{F}}|, quad ext { wrt. } operatorname{svd}(oldsymbol{F})=left[ egin{array}{lll}{sigma_{1},} & {sigma_{2},} & {0}end{array}
ight]
]
对得到的基础矩阵进行奇异值分解
[hat{oldsymbol{F}}=oldsymbol{U S V}^{T} quad ext { with } S=operatorname{diag}left(sigma_{1}, sigma_{2}, sigma_{3}
ight)
]
利用奇异值约束对基础矩阵进行重构
[oldsymbol{F}=oldsymbol{U} operatorname{diag}left(sigma_{1}, quad sigma_{2}, quad 0
ight) oldsymbol{V}^{T}
]
RANSAC-估计基础矩阵
算法流程
- 随机采样8对匹配点
- 8点法求解基础矩阵
- 奇异值约束获取基础矩阵
- 计算误差，并统计内点个数
- 重复上述过程，选择内点数最多的结果
- 对所有内点执行2,3，重新计算
内点判断标准-Sampson Distance
[egin{array}{c}{dleft(oldsymbol{x}_{1}, oldsymbol{x}_{2}
ight)=frac{left(oldsymbol{x}_{2}^{T} oldsymbol{F} oldsymbol{x}_{1}
ight)^{2}}{left(oldsymbol{F} x_{1}
ight)_{x}^{2}+left(oldsymbol{F} x_{1}
ight)_{x}^{2}+left(oldsymbol{x}_{2}^{T} oldsymbol{F}
ight)_{x}^{2}+left(oldsymbol{x}_{2}^{T} oldsymbol{F}
ight)_{y}^{2}}} \ {dleft(oldsymbol{x}_{1}, oldsymbol{x}_{2}
ight)< au}end{array}
]
RANSAC-采样次数的计算
N - 样本点个数
K - 求解模型需要最少的点的个数
p - 表示内点的概率
[K 个点都是内点概率 p^{K}
]
[K个至少有一个外点(采样失败)的概率1-p^{K}
]
[M次采样全部失败的概率left(1-p^{K}
ight)^{M}
]
[M次采样至少有1次成功的概率 quad z=1-left(1-p^{K}
ight)^{M}
]
[采样次数 quad M=frac{log (1-z)}{log left(1-p^{K}
ight)}
]
[计算p=0.9, quad K=8 ， 时，想要采样
成功率达到 z geq 0.99， 所需要的采样
次数 M
]
本征矩阵E性质
-
3x3的矩阵，秩为2
-
具有5个自由度
-
奇异值为(left[ egin{array}{lll}{sigma,} & {sigma,} & {0}end{array} ight]^{T})
求解本征矩阵
[先求解基础矩阵F
]
[widehat{oldsymbol{E}}=oldsymbol{K}_{2}^{T} oldsymbol{F} oldsymbol{K}_{1}
]
[widehat{oldsymbol{E}}=oldsymbol{U} operatorname{diag}left(sigma_{1}, quad sigma_{2}, quad 0
ight) oldsymbol{V}^{T}
]
[oldsymbol{E}=oldsymbol{U} operatorname{diag}left(frac{sigma_{1}+sigma_{2}}{2}, quad frac{sigma_{1}+sigma_{2}}{2}, quad 0
ight) oldsymbol{V}^{T}
]
相机姿态的恢复
求解Rt
[oldsymbol{E}=oldsymbol{U} Sigma oldsymbol{V}^{T}, oldsymbol{Sigma}=operatorname{diag}(sigma, quad sigma, quad 0)
]
[t_{1}=U( :, 2) quad R_{1}=U R_{Z}left(frac{pi}{2}
ight) V^{T}
]
[oldsymbol{t}_{2}=-oldsymbol{U}( :, oldsymbol{2}) quad oldsymbol{R}_{2}=oldsymbol{U} oldsymbol{R}_{mathrm{Z}}^{T}left(frac{pi}{2}
ight) oldsymbol{V}^{T}
]
[oldsymbol{R}_{z}left(frac{pi}{2}
ight)=left( egin{array}{ccc}{0,} & {-1,} & {0} \ {1,} & {0,} & {0} \ {0,} & {0,} & {1}end{array}
ight), oldsymbol{R}_{z}^{T}left(frac{pi}{2}
ight)=left( egin{array}{ccc}{0,} & {1,} & {0} \ {-1,} & {0,} & {0} \ {0,} & {0,} & {1}end{array}
ight)
]
[共有4种情况 left(oldsymbol{R}_{1}, oldsymbol{t}_{1}
ight),left(oldsymbol{R}_{1}, oldsymbol{t}_{2}
ight),left(oldsymbol{R}_{2}, oldsymbol{t}_{1}
ight),left(oldsymbol{R}_{2}, oldsymbol{t}_{2}
ight)
]
选择正确的Rt
相机的世界坐标
[oldsymbol{O}_{1}=-oldsymbol{R}^{T} oldsymbol{t}=0, quad oldsymbol{O}_{2}=-oldsymbol{R}^{T} oldsymbol{t}
]
相机的世界坐标中的朝向
[d_{1}=left[ egin{array}{lll}{0,} & {0} & {1}end{array}
ight]^{T}
]
[oldsymbol{d}_{2}=left[ egin{array}{cccc}{oldsymbol{r}_{1}^{T}} & {oldsymbol{r}_{2}^{T}} & {oldsymbol{r}_{3}^{T}} & {-oldsymbol{R}^{T} oldsymbol{t}}end{array}
ight]left[ egin{array}{l}{0} \ {0} \ {1} \ {0}end{array}
ight]=r_{3}^{T}
]
[oldsymbol{R}=left[ egin{array}{c}{oldsymbol{r}_{1}} \ {oldsymbol{r}_{2}} \ {oldsymbol{r}_{3}}end{array}
ight] quad oldsymbol{R}^{T}=left[ egin{array}{ccc}{oldsymbol{r}_{1}^{T}} & {oldsymbol{r}_{2}^{T}} & {oldsymbol{r}_{3}^{T}}end{array}
ight]
]
利用相机姿态和匹配点进行三角量测得到三维点P
P需满足同时位于两个相机的前方：
方法1：
[egin{array}{l}{left(oldsymbol{P}-oldsymbol{O}_{1}
ight)^{T} oldsymbol{d}_{1}>0} \ {left(oldsymbol{P}-oldsymbol{O}_{2}
ight)^{T} oldsymbol{d}_{1}>0}end{array}
]
方法2：
[left[ egin{array}{l}{x_{c}} \ {y_{c}} \ {z_{c}}end{array}
ight]=oldsymbol{R P}+oldsymbol{t}, quad Z_{c}>0 对两个相机成立
]
单应矩阵H
即单应矩阵有两种情况：
1.空间点位于平面
2.相机纯旋转
空间中特征点位于一平面上
[oldsymbol{n}^{T} oldsymbol{X}+d=0 quad Rightarrow quad-frac{oldsymbol{n}^{T} oldsymbol{X}}{d}=1
]
[egin{equation}
egin{aligned}
oldsymbol{x}_{2}
&= oldsymbol{K}_{2}(oldsymbol{R} oldsymbol{X}+oldsymbol{t})\
&=oldsymbol{K}_{2}left(oldsymbol{R} oldsymbol{X}+oldsymbol{t} cdotleft(-frac{oldsymbol{n}^{T} oldsymbol{X}}{d}
ight)
ight)\
&=oldsymbol{K}_{2}left(oldsymbol{R}-frac{t oldsymbol{n}^{T}}{d}
ight) oldsymbol{X} \
&=oldsymbol{K}_{2}left(oldsymbol{R}-frac{t oldsymbol{n}^{T}}{d}
ight) oldsymbol{K}_{1}^{-1} oldsymbol{x}_{1}
end{aligned}
end{equation}
]
[oldsymbol{x}_{2}=oldsymbol{H} oldsymbol{x}_{1}, quad oldsymbol{H}=oldsymbol{K}_{2}left(oldsymbol{R}-frac{oldsymbol{t} oldsymbol{n}^{T}}{d}
ight) oldsymbol{K}_{1}^{-1}
]
单应矩阵是满秩的
[x_{1}=H^{-1} x_{2}
]
t = 0时，对应纯旋转
[oldsymbol{H}=oldsymbol{K}_{2} oldsymbol{R} oldsymbol{K}_{1}^{-1}
]
单应矩阵求法
直接线性变换法
[left( egin{array}{c}{u_{2}} \ {v_{2}} \ {1}end{array}
ight)=left[ egin{array}{lll}{H_{11}} & {H_{12}} & {H_{13}} \ {H_{21}} & {H_{22}} & {H_{23}} \ {H_{31}} & {H_{32}} & {H_{33}}end{array}
ight] left( egin{array}{l}{u_{1}} \ {v_{1}} \ {1}end{array}
ight)
]
[egin{aligned}
u_{2} &=frac{H_{11} u_{1}+H_{12} v_{1}+H_{13}}{H_{31} u_{1}+H_{32} v_{1}+H_{33}} \
v_{2}& = frac{H_{21} u_{1}+H_{22} v_{1}+H_{23}}{H_{31} u_{1}+H_{32} v_{1}+H_{33}} end{aligned}
]
8个自由度，每对点有两个约束
[H_{11} u_{1}+H_{12} v_{1}+H_{13}-H_{31} u_{1} u_{2}-H_{32} u_{2} v_{1}-H_{33} u_{2}=0\
H_{21} u_{1}+H_{22} v_{1}+H_{23}-H_{31} u_{1} v_{2}-H_{32} v_{1} v_{2}-H_{33} v_{2}=0
]
令 (H_{33}=1)总共需要4对特征点
[A=left( egin{array}{c}
u_{1}^{(1)}, quad v_{1}^{(1)}, quad 1, quad 0, quad 0, quad 0, quad-u_{1}^{(1)} u_{2}^{(1)}, quad-u_{2}^{(1)} v_{1}^{(1)} \
0, quad 0, quad 0, quad u_{1}^{(1)}, quad v_{1}^{(1)}, quad 1, quad-u_{1}^{(1)} v_{2}^{(1)}, quad-v_{1}^{(1)} v_{2}^{(1)}\
{vdots} \
u_{1}^{(4)}, quad v_{1}^{(4)}, quad 1, quad 0, quad 0, quad 0, quad-u_{1}^{(4)} u_{2}^{(4)}, quad-u_{2}^{(4)} v_{1}^{(4)} \
0, quad 0, quad 0, quad u_{1}^{(1)}, quad v_{1}^{(1)}, quad 1, quad-u_{1}^{(4)} v_{2}^{(4)}, quad-v_{1}^{(4)} v_{2}^{(4)}end{array}
ight)
left( egin{array}{c}{F_{11}} \ {F_{12}} \ {F_{13}} \ {F_{21}} \ {F_{21}} \ {F_{22}} \ {F_{33}} \ {F_{31}} \ {F_{32}}end{array}
ight)
=left( egin{array}{c}{u_{2}^{(1)}} \ {v_{2}^{(1)}} \ {u_{2}^{(2)}} \ {v_{2}^{(2)}} \ {v_{2}^{(2)}} \ {u_{2}^{(3)}} \ {v_{2}^{(4)}} \ {v_{2}^{(4)}} \ {v_{2}^{(4)}}end{array}
ight)
]
RANSAC-估计单应矩阵
算法流程
- 随机采样4对匹配点(left(x_{1}^{(n)}, x_{2}^{(n)} ight))
- 8点法求解基础矩阵H
- 计算误差，并统计内点个数
- 重复上述过程，选择内点数最多的结果
- 对所有内点执行3,4，重新计算H
内点判断标准
[Eleft(x_{1}, x_{2}, H
ight)=dleft(x_{1}, H^{-1} x_{2}
ight)^{2}+dleft(x_{2}, H x_{1}
ight)^{2} \
Eleft(x_{1}, x_{2}, H
ight)< au
]