根据3.3的说法,也即是公式(14),相机的pose是IMU的pose而来。相机和IMU之间的外参是固定的,所以这两者知道一个就知道另一个了。
所谓的状态扩增就是增加一个相机的位姿,新增加的相机位姿啥都不知道? 按照文章的意思应该有: X I M U = ( q , b g , v , b a , P ) X_{IMU} = (q, b_g, v, b_a, P) XIMU=(q,bg,v,ba,P) J 6 × ( 15 + 6 N ) = ( ∂ θ ∂ X ∂ P ∂ X ) = ( ∂ θ ∂ X I M U , 0 3 × 6 N ∂ P ∂ X I M U , 0 3 × 6 N ) = ( ∂ θ ∂ q , 0 3 × 9 , ∂ θ ∂ P , 0 3 × 6 N ∂ P ∂ q , 0 3 × 9 , ∂ P ∂ P , 0 3 × 6 N ) = ( q , 0 3 × 9 , 0 3 × 3 , 0 3 × 6 N q p × , 0 3 × 9 , I 3 × 3 , 0 3 × 6 N ) J_{6\times (15+6N)}=\begin{pmatrix}\frac{\partial \theta}{\partial X}\\\frac{\partial P}{\partial X}\end{pmatrix}=\begin{pmatrix}\frac{\partial \theta}{\partial X_{IMU}},0_{3\times 6N}\\ \frac{\partial P}{\partial X_{IMU}},0_{3\times 6N}\end{pmatrix}=\begin{pmatrix}\frac{\partial \theta}{\partial q},0_{3\times 9},\frac{\partial \theta}{\partial P},0_{3\times 6N}\\ \frac{\partial P}{\partial q},0_{3\times 9},\frac{\partial P}{\partial P},0_{3\times 6N}\end{pmatrix}=\begin{pmatrix}q,0_{3\times 9},0_{3\times3},0_{3\times 6N}\\ qp\times,0_{3\times 9},I_{3\times3},0_{3\times 6N}\end{pmatrix} J6×(15+6N)=(∂X∂θ∂X∂P)=(∂XIMU∂θ,03×6N∂XIMU∂P,03×6N)=(∂q∂θ,03×9,∂P∂θ,03×6N∂q∂P,03×9,∂P∂P,03×6N)=(q,03×9,03×3,03×6Nqp×,03×9,I3×3,03×6N)
要根据IMU的位姿来计算(因为IMU是实时更新的?)。
q = q ⊗ q q=q\otimes q q=q⊗q
p = q p + p p = qp+p p=qp+p
所谓的协方差按照狄拉克记号就是 P = ∣ X > < X ∣ P=|X>
|X,x>=J|X>= ( I 15 + 6 N J 6 × ( 15 + 6 N ) ) ( 21 + 6 N ) × ( 15 + 6 N ) ∣ X 15 + 6 N > = ( X J 6 × ( 15 + 6 N ) X ) \begin{pmatrix}I_{15+6N}\\J_{6\times (15+6N)}\end{pmatrix}_{(21+6N)\times (15+6N)}|X_{15+6N}>=\begin{pmatrix}X\\J_{6\times (15+6N)}X\end{pmatrix} (I15+6NJ6×(15+6N))(21+6N)×(15+6N)∣X15+6N>=(XJ6×(15+6N)X)
继而应该有:
( θ N + 1 p N + 1 ) = J 6 × ( 15 + 6 N ) X \begin{pmatrix} \theta_{N+1}\\p_{N+1}\end{pmatrix}=J_{6\times (15+6N)}X (θN+1pN+1)=J6×(15+6N)X