Research on the Internal and External Orientation Elements of Photos Based on VC++6.0 Programming and Calculation of External Orientation Elements of Photos

According to the field survey results and aerial photos, according to the * * * line equation, using the method of single image and space intersection, the internal and external orientation elements of the photos are studied and calculated by VC++6.0 programming. Keywords: * * Linear equation VC++6.0 Single image pair space rear intersection

China Library Classification. : P23 ID number:No. A document number:1672-3791(2012) 05 (b)-0033-01In the process of aerial photogrammetry, we should do a good job in field survey and aerial photography.

Calculation principle of 1

Directional elements can be obtained by using a certain number of ground control points and the coordinates of points on the corresponding map.

Its basic idea is: based on a single image, according to the * * * line equation, solve the orientation elements of the instantaneous aerial image. Because the mathematical model * * * line equation used in space rear rendezvous is a nonlinear function, we linearize it. * * * line equation is:

After linearization, there are:

x =(x)+a 1 1 dxs+a 12 dys+a 13 dzs+a 14dφ+a 15dω+a 16dκ+a 17x+a 18y+a 19f

y =(y)+a 2 1 dxs+a 22 dys+a 23 dzs+a24dφ+a25dω+a26dκ+a27x+a28y+a29f

When we use four or more ground control points and corresponding pixel coordinates, we use the least square adjustment method to solve the problem, plus accidental errors Vx, Vy,

VX = a 1 1 dxs+a 12 dys+a 13 dzs+a 14dφ+a 15dω+a 16dκ+a 17dx+a 18dy+a 19df-Lx

vy = a 2 1 dxs+a 22 dys+a 23 dzs+a24dφ+a25dω+a26dκ+a27dx+a28dy+a29df-Ly

The coefficient of each term can be obtained by solving the partial derivative, where:

a 17 =(x-x0)/f； a27 =(y-y0)/f； a 18 = 1； a28 = 0； a 19 = 0； a29 = 1；

Then we can get the error equation:

V=AX-L

The normal equations are listed as follows: (A-T? PA)X=A-T？ PL； The solution of the normal equation is: X=(A-TA)- 1ATL.

So as to solve nine orientation elements.

2 algorithm flow chart

The calculation result of the right film is shown in figure 1.

Notepad content is exported as follows.

(1) The external orientation elements of the photo are:

xs =- 1038.485358； ys =-6 12.3039098 zs = 533.895357。

(2) The internal orientation elements of the photo are:

f = 1. 12964 1785； x0 =-4.4 14426649； y0=-5.5 1459306 .

The heading inclination is: 0.02116558532; Lateral inclination:-0.02851629385; The rotation angle of the photo is 0.02599 144.

(3) The rotation matrix r is:

0.9994657309 -0.024 1878564

-0.02 1 14962887

0.0255 173465 1 0.9992674772

0.0285 197 1209

0.02 1 15 172777 -0.029044 1573

0.9993694608

(4) The accuracy evaluation results are as follows:

The error per unit weight is (micron):

3.3878e+005 .

The calculation result of the left film is shown in Figure 2.

Notepad content is exported as follows.

(1) The external orientation elements of the photo are:

xs =- 1038.4306 17； ys =-6 12.276523 Zs = 533.6477522。

(2) The internal orientation elements of the photo are:

f =-0.7559497 103； x0 = 1.904926053； y0 =-5.557904 1 .

The heading inclination is: 0.0269046379; Lateral inclination:-0.01600371623; The rotation angle of the photo is 0.00536.

(3) The rotation matrix r is:

0.999605 192 -0.00787 1353955

-0.02697224397

0.008304493335 0.9998377524

0.0 1598449904

0.02697478945 -0.0 16202 17905

0.999508376 1

(4) The accuracy evaluation results are as follows:

The error per unit weight is (micron):

5.47 1 1e+005

3 Conclusion

According to the coordinate analysis of the image point, the midpoint of the photo is taken as the origin, the vertical axis is the Y axis, and the horizontal axis is the X axis. There is a big gap between the calculated results and the actual ones, especially the F value is negative. After using the standard data checking program, although there is a certain gap between the external orientation elements, especially xs and ys, the F value is similar. Using your own data to calculate, the deviation is too big, which becomes a mistake. Try to modify the input control point coordinate data, but there are some problems, such as the number of iterations is 1 or 29, and the size difference is too big. Therefore, the problem of data accuracy or data coordinate system difference will often bring wrong consequences to the later work.