Question 4 of 9: Deriving the Parallax Equations from Collinearity

Question 4: Deriving the Parallax Equations from Collinearity (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given / Find. Starting from the general collinearity condition, derive the vertical-photography parallax equations that give a ground point's coordinates $(X,Y,Z)$ from its measured parallax on an overlapping pair. This is a derivation, so the answer is developed as explained steps rather than a numeric computation.

1. The collinearity condition. Collinearity states that the exposure station $L$, the image point, and the ground point lie on one straight line. For a photo of focal length $f$ with exposure-station coordinates $(X_L,Y_L,Z_L)$ and rotation matrix $M=[m_{ij}]$, a ground point $(X,Y,Z)$ images at

$$x=-f\,\frac{m_{11}(X-X_L)+m_{12}(Y-Y_L)+m_{13}(Z-Z_L)}{m_{31}(X-X_L)+m_{32}(Y-Y_L)+m_{33}(Z-Z_L)},$$

with an analogous expression for $y$ (numerator using $m_{21},m_{22},m_{23}$).

2. Specialise to truly vertical photography. For a vertical photo the optical axis is plumb, so all rotation angles ($\omega,\phi,\kappa$) are zero and $M$ becomes the identity: $m_{11}=m_{22}=m_{33}=1$ and all off-diagonal terms vanish. The denominator collapses to $(Z-Z_L)$ and the equations reduce to the simple scale relations

$$x=-f\,\frac{X-X_L}{Z-Z_L},\qquad y=-f\,\frac{Y-Y_L}{Z-Z_L}.$$

3. Insert the aerial sign/height convention. The camera is above the ground, so $Z_L=H$ (flying height) and $Z=Z_A$ (ground elevation); writing the positive downward object distance as $H-Z$ absorbs the sign, giving the familiar vertical-photo forms

$$x=f\,\frac{X-X_L}{H-Z},\qquad y=f\,\frac{Y-Y_L}{H-Z}.$$

4. Write both exposures of the stereo-pair. Let the left exposure station be the coordinate origin, $L_1=(0,0,H)$, and the right station be displaced by the air base $B$ along the flight ($X$) axis, $L_2=(B,0,H)$ (a vertical pair shares the same $H$ and $Y_L$). The same ground point images on each photo as

$$x=f\,\frac{X}{H-Z}\quad(\text{left}),\qquad x'=f\,\frac{X-B}{H-Z}\quad(\text{right}).$$

5. Difference the two to introduce parallax. The $x$-parallax is $p=x-x'$; subtracting the right relation from the left, the $X$ terms cancel and only the base survives:

$$p=x-x'=f\,\frac{X-(X-B)}{H-Z}=\frac{Bf}{H-Z}.$$

Rearranging gives the first parallax equation, the parallax–height relation:

$$\boxed{\,H-Z=\frac{Bf}{p}\quad\Longleftrightarrow\quad Z=H-\frac{Bf}{p}\,}.$$

6. Back-substitute for the planimetric coordinates. From the left-photo relation $X=\dfrac{x\,(H-Z)}{f}$, replace $(H-Z)$ using the result just derived, $(H-Z)=Bf/p$:

$$X=\frac{x}{f}\cdot\frac{Bf}{p}=\frac{B\,x}{p},\qquad\text{and identically}\qquad Y=\frac{B\,y}{p}.$$

Collecting the three results gives the complete set of parallax equations (model coordinates referred to the left exposure station):

$$\boxed{\;X=\dfrac{B\,x}{p},\qquad Y=\dfrac{B\,y}{p},\qquad Z=H-\dfrac{Bf}{p}\;}$$

They are nothing more than the collinearity equations stripped of rotation (vertical photo) and then differenced across the base — which is why they only hold for near-vertical photography and degrade as the tilts grow. In practice small residual tilts are removed first (rectification / relative orientation) so that the parallax equations may be applied.