Question 2 of 9: Elevations from a Vertical Stereo-pair (Parallax)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2014 — 3 hours, closed book (any non-communicating calculator permitted). SEVEN questions constitute a complete paper: Part A answer all of #1–#5, Part B answer one of #6/#7, Part C answer one of #8/#9. Marks are shown in brackets. All nine questions (including both alternatives in Parts B and C) are solved below for completeness.
Reference texts: Wolf, Dewitt & Wilkinson, Elements of Photogrammetry with Applications in GIS (4th ed., McGraw-Hill, 2014); Mikhail, Bethel & McGlone, Introduction to Modern Photogrammetry (Wiley, 2001); Kraus, Photogrammetry: Geometry from Images and Laser Scans (2nd ed., de Gruyter, 2007); Ghilani & Wolf, Elementary Surveying (15th ed.). Canadian mapping practice (NRCan / Canadian Geodetic Survey) throughout.
Question 2: Elevations from a Vertical Stereo-pair (Parallax) (15 marks)
Given. A vertical stereo-pair with the flight line along the $x$-axis, flying height $H=1230$ m above MSL, and measured $x$-parallaxes on three points; point C (on the flight line) has a known elevation used to calibrate the unknown air-base × focal-length product.
Point
$x$ (left)
$x'$ (right)
parallax $p=x-x'$
A
$53.41$
$-38.26$
$91.67$ mm
B
$88.92$
$-7.06$
$95.98$ mm
C (known $h_C=590$ m)
$14.3$
$-78.3$
$92.6$ mm
Find. The elevations (above MSL) of points A and B.
Vertical stereo-pair: a higher ground point (smaller $H-h$) yields a larger $x$-parallax $p=x-x'$. With $p=Bf/(H-h)$, one point of known elevation (C) fixes the constant $Bf$, after which every other point's elevation follows from its parallax.
Approach. For a vertical pair the parallax–height equation is $p=Bf/(H-h)$; use the known point C to evaluate the constant $Bf$, then invert for the elevations of A and B.
Form the $x$-parallaxes. Parallax is the algebraic difference of the $x$-photo-coordinates (flight-axis parallel to $x$):
$$p_A = 53.41-(-38.26)=91.67\ \text{mm},\quad p_B = 88.92-(-7.06)=95.98\ \text{mm},$$
$$p_C = 14.3-(-78.3)=92.6\ \text{mm}.$$
Calibrate the constant $Bf$ from point C. Rearranging $p=Bf/(H-h)$ gives $Bf=(H-h)\,p$. Neither $B$ nor $f$ is given, but their product is fixed by C:
$$Bf = (H-h_C)\,p_C = (1230-590)(92.6)=640\times92.6=\boxed{59\,264\ \text{m}\cdot\text{mm}}.$$
Elevation of A. Invert the parallax equation, $h = H - Bf/p$:
$$h_A = 1230 - \frac{59\,264}{91.67} = 1230 - 646.5 = \boxed{583.5\ \text{m (MSL)}}.$$
Elevation of B. Same relation with $p_B$:
$$h_B = 1230 - \frac{59\,264}{95.98} = 1230 - 617.5 = \boxed{612.5\ \text{m (MSL)}}.$$
The $y$-coordinates are not needed for elevation here: with the flight line along $x$, only the $x$-parallax carries the height information. Point A, with the smaller parallax, sits lower than B, exactly as the geometry requires.
