lidar (laser scanner) georeferencingf (sensor model)

Derivation of projective transformation based on shift, rotation and scale parameters

Reduced normal matrix for solving fast photogrammetric bundle adjustment

Reduced normatrix for photogrammetric bundle adjustment based on the collinearity equations

Note: There are $m_1$ photos and $m_2$ object points, and the number of EOPs is 6. If 6*$m_1$ is less than 3*$m_2$, solve unknowns related with EOPs. If not, solve object points coordinates, first. This post shows only the first case, 3*$m_2$ > 6*$m_1$.

[Fig. Normal matrix sample]

$$ \begin{pmatrix} N_1 & N_{12} \\ N_{12}^T & N_2 \end{pmatrix}\begin{pmatrix} X_1 \\ X_2 \end{pmatrix} = \begin{pmatrix} C_1 \\ C_2 \end{pmatrix} $$ $$ N_1 X_1 + N_{12} X_2 = C_1$$ $$ N_{12}^T X_1 + N_2 X_2 = C_2$$ In case $6m_1<3m_2$, let's slove $X_1$ first.
$$ X_2 = N_2^{-1} C_2 - N_2^{-1} N_{12}^T X_1$$, therefore
$$ (N_1 - N_{12} N_2^{-1} N_{12}^T) X_1 = C_1 - N_{12} N_2^{-1} C_2$$ $$ X_1 = (N_1 - N_{12} N_2^{-1} N_{12}^T)^{-1}(C_1 - N_{12} N_2^{-1} C_2)$$ And
$$ X_2 = N_2^{-1}(C_2 - N_{12}^T X_1) $$ For $X_2$, we can avoid solving the inverse matrix of $3m \times 3m$ .
For $i_{th}$ point,
$$ X_{2_i} = N_{2_{i}}^{-1}(C_{2_i} - N_{12_i}^T X_1) $$

Photogrammetric Intersection

Linear Intersection for a Stereo Pair

There are two images.
Object point vs. image point, in the first image:
$$ \begin{pmatrix} X \\ Y \\ Z \end{pmatrix} = \begin{pmatrix} X^0_a \\ Y^0_a \\ Z^0_a \end{pmatrix} + {\lambda_a}{R_a} \begin{pmatrix} x_a \\ y_a \\ -f_a \end{pmatrix} \tag{1} $$
Object point vs. image point, in the second image:
$$ \begin{pmatrix} X \\ Y \\ Z \end{pmatrix} = \begin{pmatrix} X^0_b \\ Y^0_b \\ Z^0_b \end{pmatrix} + {\lambda_b}{R_b} \begin{pmatrix} x_b \\ y_b \\ -f_b \end{pmatrix} \tag{2} $$
Therefore,
$$ \begin{pmatrix} X^0_b \\ Y^0_b \\ Z^0_b \end{pmatrix} - \begin{pmatrix} X^0_a \\ Y^0_a \\ Z^0_a \end{pmatrix} = {\lambda_a}{R_a} \begin{pmatrix} x_a \\ y_a \\ -f_a \end{pmatrix} - {\lambda_b}{R_b} \begin{pmatrix} x_b \\ y_b \\ -f_b \end{pmatrix} \tag{3} $$ $$ \begin{pmatrix} r_{11_a}x_a + r_{12_a}y_a - r_{13_a}f_a & - r_{11_b}x_b - r_{12_b}y_b + r_{13_b}f_b \\ r_{21_a}x_a + r_{22_a}y_a - r_{23_a}f_a & - r_{21_b}x_b - r_{22_b}y_b + r_{23_b}f_b \\ r_{31_a}x_a + r_{32_a}y_a - r_{33_a}f_a & - r_{31_b}x_b - r_{32_b}y_b + r_{33_b}f_b \end{pmatrix} = \begin{pmatrix} X^0_b - X^0_a \\ Y^0_b - Y^0_a \\ Z^0_b - Z^0_a \end{pmatrix} \tag{4} $$
where, $ \begin{pmatrix} X \ Y \ Z \end{pmatrix}^T $ are object coordinates,
$\begin{pmatrix} X^0_a \ Y^0_a \ Z^0_a \end{pmatrix}^T$, $\begin{pmatrix} X^0_b \ Y^0_b \ Z^0_b \end{pmatrix}^T$, $R_a$, and $R_b$ denote positoions and attitudes of two images. $\begin{pmatrix} x_a \ y_a \ -f_a \end{pmatrix}^T$ and $\begin{pmatrix} x_b \ y_b \ -f_b \end{pmatrix}^T$ are photo coordinates of each image. If two images are obtained by a camera, $f_a = f_b$.
In the equation (4), there are two unknowns ($\lambda_a$ and $\lambda_b$) and three linear equations. Once $\lambda_a$ and $\lambda_b$ are determined, we get two sets of object coordinates from two scale factors and can use mean values of them.

General Case: Linear Intersection for Multiple Images

$$ \left. \begin {matrix} \begin{pmatrix} X \\ Y \\ Z \end{pmatrix} - {\lambda_1}{R_1} \begin{pmatrix} x_1 \\ y_1 \\ -f_1 \end{pmatrix} = \begin{pmatrix} X^0_1 \\ Y^0_1 \\ Z^0_1 \end{pmatrix} \\ \begin{pmatrix} X \\ Y \\ Z \end{pmatrix} - {\lambda_2}{R_2} \begin{pmatrix} x_2 \\ y_2 \\ -f_2 \end{pmatrix} = \begin{pmatrix} X^0_2 \\ Y^0_2 \\ Z^0_2 \end{pmatrix} \\ \vdots \\ \begin{pmatrix} X \\ Y \\ Z \end{pmatrix} - {\lambda_n}{R_n} \begin{pmatrix} x_n \\ y_n \\ -f_n \end{pmatrix} = \begin{pmatrix} X^0_n \\ Y^0_n \\ Z^0_n \end{pmatrix} \end {matrix} \right ] \tag{5} $$
Threrefore,
$$ \begin{pmatrix} 1 & 0 & 0 & -r_{11_1}x_1 - r_{12_1}y_1 + r_{13_1}f_1 & \cdots \\ 0 & 1 & 0 & -r_{21_1}x_1 - r_{22_1}y_1 + r_{23_1}f_1 & \cdots \\ 1 & 0 & 1 & -r_{31_1}x_1 - r_{32_1}y_1 + r_{33_1}f_1 & \cdots \\ 1 & 0 & 0 & 0 & -r_{11_2}x_2 - r_{12_1}y_2 + r_{13_2}f_2 & \cdots \\ 0 & 1 & 0 & 0 & -r_{21_2}x_2 - r_{22_2}y_2 + r_{23_2}f_2 & \cdots \\ 0 & 0 & 1 & 0 & -r_{31_2}x_2 - r_{32_2}y_2 + r_{33_2}f_2 & \cdots \\ \vdots \\ 1 & 0 & 0 & 0 & 0 & \cdots & -r_{11_n}x_n - r_{12_n}y_n + r_{13_n}f_n \\ 0 & 1 & 0 & 0 & 0 & \cdots & -r_{21_n}x_n - r_{22_n}y_n + r_{23_n}f_n \\ 0 & 0 & 1 & 0 & 0 & \cdots & -r_{31_n}x_n - r_{32_n}y_n + r_{33_n}f_n \end{pmatrix} \begin{pmatrix} X \\ Y \\ Z \\ \lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n \end{pmatrix} = \begin{pmatrix} X^0_1 \\ Y^0_1 \\ Z^0_1 \\ X^0_2 \\ Y^0_2 \\ Z^0_2 \\ \lambda_2 \\ \vdots \\ X^0_n \\ Y^0_n \\ Z^0_n \end{pmatrix} \tag{6} $$
In the equation (6), there are $(3+n)$ unknowns ($\begin{pmatrix}X&Y&Z\end{pmatrix}^T$, and $3n$ equations. If there are 2 or more than 2 images, there should be redundancy.

Derivation of direct linear transformation for a line scanner

: focal length
: rotation matrix element
: CCD array coordinate (column)
: offset (for the three line scanner)

: First order function of scan line
: Constant (zero order function of scan line)
: Scan line number


 




-. Matrix form

Plane fitting

 

Plane equation:

(1) lx + my + nz =  d

where d is a distance from an origin to the plane,

{l,m,n} is direction consines, and

sqrt(l*l + m*m + n*n)=1.

(2) ax + by + cz = 1

where d presented in (1) can be calculated by 1/sqrt(a*a + b*b + c*c).

(3) Ax + By + Cz = d^2

where (A, B, C) denotes a vector from an origin to the plane, which can be obtained by A = l*d, B=m*d, and C=n*d.

 

For a point cloud, P0, P1, P2, ~ Pn,

Its centroid C, (cx, cy, cz), is on an expected least squre fitted plane.

Therefore, dot product of [l, m, n] and (Pi-C) is zero.

l*xi + m*yi + n*zi = 0

whrer (xi,yi,zi) = Pi

To avoid [l, m, n] are estimated as zero,

a constraint of direction consines is adopted,

l^2 + m^2 + n^2 = 1.0