lidar (laser scanner) georeferencingf (sensor model)

Multi-byte vs. Unicode

Multi-byte	Unicode
char	TCHAR
strcat_s()	_tcscaft_s()
strcpy_s()	_tcscpy_s()
strncpy_s()	_tcsncpy_s()
strien()	_tcsien()
sprinif_s()	_stprintf_s()

How to get paths defined in the system environment

char path[_MAX_PATH];

const char* varName = "[system environment variable name]";
size_t len;
getenv_s(&len, path, 80, varName);

Print out a (decimal) number as a hexadecimal number

int decNum = 123456;
char hexNum[16];

// decimal number to hexadecimal number
sprintf_s(hexNum, 16, "%X", decNum);
printf("%d -> %s \n", decNum, hexNum);

// hexadecimal number to decimal number
sprintf_s(hexNum, 16, "FF");
decNum = (int)strtol(hexNum, nullptr, 16);
printf(" %s -> %d \n", hexNum, decNum);

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,
$$ X_{2_i} = N_{2_{i}}^{-1}(C_{2_i} - N_{12_i}^T X_1) $$

Brief memo for the units in least squares

-. A-priori reference variable has no unit (unitless). It is typically 1.0 or can be a different value for scaling.
-. Weight is determined by a-prior reference variable and a-prior variance of each observation. Therefore, the unit of weight is the same as the square of the observation unit.
-. A-posterior reference variance has no unit.
-. The variances of estimated unknown parameters after adjustment are the same as the square of the unknown parameters unit.

How to calculate areas of triangles and polygons

3 vertices and cross product

If 3 vertices' coordinates are given, the size of the cross product of the two vectors is same to two times of the triangle area.
삼각형의 3꼭지점 좌표를 알고 있다면, 인접한 두 벡터의 외적의 크기는 삼각형 면적의 두배와 같다.

$$ A = 0.5|(p_2-p_1)\times(p_3-p_1)| \tag{1}$$

If $\vec{N}$ is the result of the cross product of the two vectors, $\vec{p_1p_2}$ and $\vec{p_1p_3}$, the size of $\vec{N}$ is same to the parallelogram including $\vec{p_1p_2}$ and $\vec{p_1p_3}$.
만약 $\vec{N}$이 두 벡터 $\vec{p_1p_2}$ and $\vec{p_1p_3}$의 외적이면, $\vec{N}$의 크기는 두 벡터 두 벡터 $\vec{p_1p_2}$ and $\vec{p_1p_3}$을 포함하는 평행사변형의 면적과 같다.

Using the cross product and vertices' coordinates, we can get the area of a polygon by spliting a polygon into sub-triangles.
벡터외적과 꼭지점들의 좌표를 이용하면, 다각형을 내부의 작은 삼각형으로 분할하여 그 면적을 계산할 수 있다.

$$ A = \sum_{i=2}^{n-1} 0.5|(p_i-p_1)\times(p_{i+1}-p_1)| \tag{2}$$

We should know the limitation that any vector defiend by two vertices cannot intersect with any side of the polygon.
두 개의 꼭지점으로 정의되는 어떤 벡터도 다각형의 어떤 변과도 만나면 안된다는 한계점을 알고 있어야 한다.

This page will be update for how to calculate a complex polygon's area.

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.

[GoodToKnow] MathJax for Blogger

Aftter adding this line,
<script type="text/x-mathjax-config"> MathJax.Hub.Config({ tex2jax: { inlineMath: [ ['$','$'], ["\\(","\\)"] ], processEscapes: true } }); </script> <script src='https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.2/MathJax.js?config=TeX-MML-AM_CHTML'></script>
in 'head' of the blogger template editor in html mode, you can use MathJax expressions.
For a new line, use this block,

$$ ... $$
and, for an inline equation, use this line,
$ ... $