Photography

Black & white (panchromatic) aerial photos (23 ´ 23 cm) from Hyytiälä in Central Southern Finland have been scanned at 21 µm resolution. The original scale of the vertical photos is approx. 1:10000. In other words the pixel width "on ground" is 21 centimeters.

Orientation

The inner and exterior orientation of the images have been solved enabling 3D stereo restitution at approx. 15 centimeters accuracy.

The solution of the inner orientation involves three + four parameters

- the focal length of the pinhole camera (camera constant, distance of the projection center and the film plane)

- the location (x,y on the film) of the principal point (the point on the film plane which lies normal/perpendicular to the projection center)

When the aerial photo is scanned the coordinate system is changed to the pixel coordinate system of the digital photo. A four parameter transformation (offset in x, offset in y, rotation, scale) is used to transform pixel coordinates into film (camera) coordinates and back.

Exterior orientation means that we know the location and attitude of the camera (projection center and film plane). The location of the camera is given by three coordinates (X₀, Y₀, Z₀) and the attitude is defined with three rotation angles (rotation about the X-, Y-, and Z-axis).

Hence there are six parameters that establish the exterior orientation.

Equations in (1, above) are called the collinearity equations. Collinearity equations map the 3D real world objects (X,Y,Z) to 2D image observations (x,y). Parameters of inner orientation are (x_o,y_o, c). The three angle parameters of the exterior orientation are implicitly defined in the coefficients r_ij and the location of the projection center are given by the terms (X₀, Y₀, Z₀).

By looking at equation 1 it is clear that one image is not sufficient for 3D-reconstruction of objects from their perspective projections! There are namely too many (3) unknowns (X,Y,Z) to be solved from two equations!

3D from orientated photos

Adding another image renders two more equations (in eq. 1) and with four collinearity equations it is possible to solve three unknown coordinates!
 
 

Picture 2. With two or more images available it is possible to calculate the location of the intersecting rays that projected from the target onto the film(s). In the example we are interested in the 3-D location of point P. If we see the tree top in the left image we should be able to solve the correspondence problem for the right (others) image. I.e. we should be able to locate the same entity on all image involved (this is also called matching).

A demo program kuvamitt.exe

Picture 3. kuvamitt.exe demonstrates the use of 2 or 3 images for stereo restitution. The images are however not viewed in stereo! The user solves the correspondence problem i.e. points and clicks the same target on all images.

Picture 4. The camera coordinates (for the points that have been measured, clicked) appear in text boxes on the left. There are four combinations (in the case of an image triplet) to choose from. The check boxes are used to select which image observations are used in estimating the 3D-coordinates. The button "Calculate space intersection i.e. compute X, Y, Z from image observations" solves the ray intersection. The results (X,Y,Z) appear if the solution is found. "Demo VLL" button demonstrates Vertical line locus method. The "Demo epipolar line" -button shows the epipolar line on the middle image based on the image observation on the left image. If we click a point on the left image the matched entity is to be found in a search space on the middle image which reduces into a line (that's called epipolar line, epipolar constraint). The initial approximations for X,Y,Z text boxes shown the initial approximations for the unknown coordinates. Ray intersection is an unlinear problem and hence an iterative (gradient based) method is used and it requires that an approximation is given. The coordinates are from the Finnish KKJ-system (reverse Mercator projection)

Picture 5. There are six image pairs and three image triplets to choose from. The names are familiar to forestry students who've been to Hyytiälä.
 
 

Exercise

Image pair photos

SMEAR tower images

What is the elevation of the tower top?
Is the tower tilted or does it stand vertical?

"A hill"

What is the elevation of the highest point? Where is it in KKJ-east and north ?

Hyytiälä building images

* There are two buildings in the low right corner of the images.

How high are the buildings?
What is the volume of the bigger building in m3?

* The uppermost parts of the image show a lawn field.

What is the elevation of the field (it is flat). Look at column 2746, row 10843 on the left image?

Spruce plot 1

* What is the terrain elevation in the forested part of the photo? Look for shadows cast on the ground.
* How high are the trees? Click a tree top on the left image and use the epipolar line to find the corresponding tree top on the right hand side image.

"Clearing image"

* there are two shadows of trees in the middle parts. What is the height of the Trees based on the shadow lengths ? (elevation of the sun is 34 degrees above the horizon, the azimuth angle is 115 degrees from KKJ-north)

Kalela's stand

* what is the elevation of the log pile visible in the middle of the image?
* how long are the logs?

Image triplets

"Konehalli images"

* measure the elevation of the roof of the buildings using different combinations of images: 1+2, 1+3, 2+3, 1+2+3