ENGI 3703 Surveying and Geomaticssitotaw/Site/Fall2007_files/Lecture23.pdfENGI 3703 Surveying and...
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ENGI 3703 Surveying and Geomatics Topic Instructor: Prof. Ken Snelgrove Lect 23 - Nov 14/07 Slide 1 of 6 Photogrammetry - Parallax Parallax Measurement : (Section 27.10) We have seen that by rules of simple geometry that we can determine the elevation of vertical objects in a vertical air photo. This is due to the fact that the base of the object and top are both visible in the photo due to the effect of relief displacement. Equations developed from similar triangles allow the elevation to be determined because the top and bottom of the object share a common distance from the “principal point”. We can also determine the difference in height of objects that are distant from one another. However, to do this we require more information from a second air photo and use of the “parallax” phenomena. Parallax is the apparent displacement of objects due the changing location of the observer. Parallax Demonstration - Hold up a finger at arms length, close one eye and focus on a distant object. Keeping on eye closed and your finger stationary, turn your head from side to side. You’ll notice that your finger apparently moves a greater distance than the distant object. This is parallax! In aerial photography the camera moves with the aircraft and images that are closer (higher elevation objects) appear to farther than more distant ones (lower elevation). If we have overlapping aerial photos, we can use this knowledge to extract elevation data from aerial photographs.
Transcript of ENGI 3703 Surveying and Geomaticssitotaw/Site/Fall2007_files/Lecture23.pdfENGI 3703 Surveying and...
ENGI 3703Surveying and Geomatics
TopicInstructor: Prof. Ken Snelgrove
Lect 23 - Nov 14/07 Slide 1 of 6Photogrammetry - Parallax
Parallax Measurement: (Section 27.10) We have seen that by rules of simple geometry that we can
determine the elevation of vertical objects in a vertical air photo. This is due to the fact that the base of
the object and top are both visible in the photo due to the effect of relief displacement. Equations
developed from similar triangles allow the elevation to be determined because the top and bottom of
the object share a common distance from the “principal point”.
We can also determine the difference in height of objects that are distant from one another. However,
to do this we require more information from a second air photo and use of the “parallax” phenomena.
Parallax is the apparent displacement of objects due the changing location of the observer.
Parallax Demonstration - Hold up a finger at arms length, close one eye and focus on a distant object.
Keeping on eye closed and your finger stationary, turn your head from side to side. You’ll notice that
your finger apparently moves a greater distance than the distant object. This is parallax!
In aerial photography the camera moves with the aircraft and images that are closer (higher elevation
objects) appear to farther than more distant ones (lower elevation). If we have overlapping aerial
photos, we can use this knowledge to extract elevation data from aerial photographs.
ENGI 3703Surveying and Geomatics
TopicInstructor: Prof. Ken Snelgrove
Lect 23 - Nov 14/07 Slide 2 of 6Photogrammetry - Parallax
Aerial Photo Image Pairs - St. John’s, NL July 1995 (National Air Photo Library)
Direction of FlightDirection of Flight
Principal
PointPrincipal
Point 1
Corresponding
Principal Point 1Corresponding
Principal Point
Y-axisY-axis
Left Image Right Image
X+X- X+X-
ENGI 3703Surveying and Geomatics
TopicInstructor: Prof. Ken Snelgrove
Lect 23 - Nov 14/07 Slide 3 of 6Photogrammetry - Parallax
Parallax Equations
Difference in x between a & a1 is the parallax measure (p)
p = x x1
O,O1
L,L1
A1’ A’
B
a1’ a’o,o1
pH-h
f
Here we have a common principal axis
and can develop similar triangles LA1’A’
and La1’a’
p
f=
B
H h
Solving for H-h
H h =Bf
p
x
X
Also we have triangles LOA’
and Loa’
x
X=
f
H h
Solving for H-h
H h =Xf
x
Equating both
B f
p= H h =
X f
x
X =B
px similarly Y =
B
py
ENGI 3703Surveying and Geomatics
TopicInstructor: Prof. Ken Snelgrove
Lect 23 - Nov 14/07 Slide 4 of 6Photogrammetry - Parallax
Parallax Example - Find the Elevation of St John’s Harbour and Windsor Lake
Direction of FlightDirection of Flight
Corresponding
Principal Point
Y-axis Y-axis
Left Image Measurements Right Image
Xa
Xb Xb1
Xa1
xa=4.28 in
xb=2.53 in
b =3.63 in
xa1=0.80 in
xb1=-1.05 in
ya1=3.85 in
yb1=-2.91 in
pi=xi-xi1
pa=4.28-0.80 = 3.48 in
pb=2.53-(-1.05)= 3.58 in
ya1
yb1
b
Parallax Measurements
ENGI 3703Surveying and Geomatics
TopicInstructor: Prof. Ken Snelgrove
Lect 23 - Nov 14/07 Slide 5 of 6Photogrammetry - Parallax
Find Air Base (B):we must find the distance from the Principal Point (PP) to the Corresponding Principal Point (CPP) on the air
photo (b). This distance must be adjusted for relief displacement and must be used to determine the flying height
(H) at the same time. For this we can utilize a topographic map to determine the airbase (B) and from this extract
the flying height (H).
We measure B from a topographic map as 9500 ft and the elevation of the CPP is hB = 670 ft (you would be
given this [see map next page]). We can determine the flying height (H) if we know the the elevation of the
Corresponding Principal Point from:
p
f=
B
H h
h = HBf
p
h =163729500(6)
3.48h = 7 ft
L2=
(H hB )xbf
(H hA )xaf
2
+(H hB )yb
f
(H hA )yaf
2
along the principal axis all y measurements are zero. In addition,
the x measurement at the principal point is zero this leaves:
L2=
(H hB )xbf
2
95002=
(H 670.0)3.636
2
H =16372 ftp
f=
B
H h
h = HBf
p
h =163729500(6)
3.58h = 450 ft
Elevation of Harbour (actual 0 ft)
Elevation of Windsor Lake (actual 460 ft)
ENGI 3703Surveying and Geomatics
TopicInstructor: Prof. Ken Snelgrove
Lect 23 - Nov 14/07 Slide 6 of 6Photogrammetry - Parallax
PP
CPP
1 km
