Question 3 of 9: Terrestrial Stereometric Camera (Normal Case)
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 3: Terrestrial Stereometric Camera (Normal Case) (15 marks)
Given. A terrestrial (horizontal) stereometric normal-case setup:
Quantity
Value
Camera base $b$ (parallel to $X$, left→right positive)
$1.20$ m
Focal length $f$
$64$ mm
Left perspective centre $X_L,\,Y_L,\,Z_L$
$100,\,100,\,100$ m
Left photo: $x_a,y_a$ / $x_b,y_b$
$31.40, 23.75$ / $5.10, -4.25$ mm
Right photo: $x'_a$ / $x'_b$
$28.00$ / $-8.25$ mm
Find. (a) which of a, b is closer to the camera base; (b) the slope distance between the two ground points A and B.
Normal-case terrestrial pair, plan view: optical axes perpendicular to the base. Depth from the base is $Y=bf/p$; the larger parallax of b ($p_b=13.35$ mm vs $p_a=3.40$ mm) means b lies much closer to the camera base than a.
Approach. In the normal case the ground coordinates follow directly from parallax: depth $Y=bf/p$, and $X=X_L+b\,x/p$, $Z=Z_L+b\,y/p$. Compute both points, compare depths for (a), then take the 3-D distance for (b).
Parallaxes. With the base along $X$, $p=x-x'$:
$$p_a = 31.40-28.00 = 3.40\ \text{mm},\qquad p_b = 5.10-(-8.25)=13.35\ \text{mm}.$$
Depths from the base — part (a). Depth is $Y=bf/p$:
$$Y_a=\frac{1.20\times64}{3.40}=22.59\ \text{m},\qquad Y_b=\frac{1.20\times64}{13.35}=\boxed{5.75\ \text{m}}.$$
Since $Y_bpoint b is the closer point to the camera base.
Full ground coordinates. Using $X=X_L+b\,x/p$ and $Z=Z_L+b\,y/p$ (and $Y=Y_L+bf/p$ measured from the base):
$$A:\ X_A=100+\tfrac{1.20(31.40)}{3.40}=111.08,\ \ Y_A=100+22.59=122.59,\ \ Z_A=100+\tfrac{1.20(23.75)}{3.40}=108.38\ \text{m},$$
$$B:\ X_B=100+\tfrac{1.20(5.10)}{13.35}=100.46,\ \ Y_B=100+5.75=105.75,\ \ Z_B=100+\tfrac{1.20(-4.25)}{13.35}=99.62\ \text{m}.$$
Slope distance AB — part (b). The 3-D Euclidean distance:
$$\Delta X=10.62,\ \Delta Y=16.84,\ \Delta Z=8.76\ \text{m},$$
$$D_{AB}=\sqrt{10.62^{2}+16.84^{2}+8.76^{2}}=\boxed{21.75\ \text{m}}.$$
The $Z$-values confirm the geometry: A is above the perspective-centre height (its $y_a$ is positive) while B is slightly below it, and b's short depth dominates the coordinate differences.
