Distortion and Displacement
Distortion – “…any shift in the position of an image on a photograph that alters the perspective
characteristics of the image.”
Displacement – “…any shift in the position of an image on a photograph that does not alter the
perspective characteristics of the photograph”
Types of Distortion
Types of Displacement
Film & print shrinkage
Curvature of the Earth
Atmospheric refraction of light Image motion (incl. object ht.)
Lens distortions
Tilt
Topography or relief
Lens distortion radiates from the PP
• Causes an image to appear closer to or farther from the PP
• most serious near the edges of the photo
Usually
negligible
Distortion and Displacement
I.
Principal Points (PP) & Conjugate Principal Points (CPP) – Not the only points of interest on
an aerial photo
A)
Nadir & Isocenter – also important, especially on oblique photos.
B)
PP – physical or optical center of an uncropped photo (intersection of diagonal fiducials)
1) If a photo is truly vertical, PP is directly beneath the aircraft
2) Not so on and obliques
3) Distortions due to lens imperfections radiate from the PP, but are typically very
slight….so they are ignored.
C)
Nadir is the gravitational center of the photo, and is directly beneath the plane
regardless of flight attitude (pitch & roll) at the time of film exposure.
1) If tilt is not too great, nadir will be on the photo
2) Displacement due to elevation differences are radial from Nadir
3) Displacement do to elevation differences can cause serious problems when trying to
measure distance and directions accurately
4) Displacement do to elevation differences is what allows us to see photos in stereo
and will provide us with 2 of the 3 techniques for heights from aerial photos.
Distortion and Displacement
Extreme effects of topographic displacement and distortion…
•
•
Can we accurately measure scale, distance, or area using this photo?
Is average PRS helpful for area calculation?
Displacement
However, Terrain Variation…
• displaces both the top & bottom of
vertical objects.
• How much & direction (toward/away
from Nadir) depends on magnitude of
elev. ∆ (+/-) w/respect to Nadir elev.
On a vertical photograph over flat terrain…
• The farther any vertical object is from
nadir…the greater the object’s top is
displaced away from nadir
• The object’s bottom position remains
stationary (i.e., no displacement)
Distortion and Displacement
D) Isocenter – is always on a line
through the PP & Nadir
called the Principal Meridian,
and is located half way
between these two points.
1) Displacement due to tilt is radial from Isocenter
2) Displacement due to tilt is zero on a truly vertical photos, but increases proportionally as
tilt angle increases.
3) Tilt displacement on slightly tilted photos is usually less in magnitude than displacement
from elevation differences, but tilt is much more difficult to detect, calculate, and correct.
(easier to detect in Chicago vs. rural Iowa…why?)
•
Images on “up side” of a tilted photo are
displaced toward photo center
•
Images on “down side” of titled photo
displace away from photo center
Topographic Displacement
For photo scale problems, we have used these symbols:
D = Ground distance  GD
d = Photo distance  PD
Hg = Flying height above ground  H
From now on used these symbols
GD = Ground distance
PD = Photo distance
H = Flying height above ground/datum
Why? Due to symbology overlap with displacement problems.
H
h
d
r
A
E
=
=
=
=
=
=
Flying height above ground
Height (+/–) of an object on the ground (tree, building, etc.)
Displacement distance on photo (inches or mm)
Radial distance on photo (inches or mm) from nadir to the displaced point
Altitude of aircraft above MSL
Elevation of ground level or datum
Displacement:
All about Similar Triangles
In this Example…
r = photo dist.: nadir to building top
r’ = photo dist.: nadir to building base
d = displacement dist.: top – base  r – r’
Displacement ( d )
d = r – r’
Consider following relationships:
f  r
as H-h  R
f  r’ as H  R
Since d = r – r’
We have to solve for
r and r’ separately…
MSL
Topographic Displacement:
1
𝑓
𝐻−ℎ
=
𝑟
𝑅
∴
𝑓(𝑅)
𝑟=
𝐻−ℎ
All about Similar Triangles
𝑑 = 𝑟 − 𝑟′
𝑓𝑅
𝑑 = 𝐻−ℎ −
𝑓𝑅
𝐻
𝐻(𝑓𝑅)
𝑑 = 𝐻(𝐻−ℎ) −
2
𝑓
𝐻
=
𝑟′
𝑅
∴
𝑓(𝑅)
𝑟′ =
𝐻
𝐻−ℎ 𝑓𝑅
𝐻 𝐻−ℎ
𝑑=
𝐻(𝑓𝑅)
𝐻(𝐻−ℎ)
𝑑=
𝐻𝑓𝑅 − 𝐻𝑓𝑅 + ℎ𝑓𝑅
𝐻(𝐻−ℎ)
ℎ𝑓𝑅
𝑑 = 𝐻(𝐻−ℎ)
𝑑=
𝑟ℎ
𝐻
−
𝐻𝑓𝑅−ℎ𝑓𝑅
𝐻 ℎ−𝐻
r
𝑑 = 𝑟 − 𝑟′
Topographic Displacement at ISU
Measurements
based on…
Photo # 30
Scale 1:24100
H = 12000’
NADIR
Suburb
E = 1035.07’
hsub = +119.09’
River
𝑫𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕: 𝒅 =
Actual
Actual
𝑑𝑟𝑖𝑣
E = 876.05’
hriv = -39.93’
𝒓𝒉
𝑯
rriv  nadir to river  13.15 * 0.2 = 2.63”
rsub  nadir to suburb  18.65 * 0.2 = 3.73”
𝑟𝑟𝑖𝑣 ∗ ℎ𝑟𝑖𝑣
2.63" ∗ (−39.93′)
=
=
= −0.00875"
𝐻𝑛𝑎𝑑𝑖𝑟
12000′
𝑑𝑠𝑢𝑏
Displacement
NADIR
E = 915.98’
Correct map
position 0.00875”
farther away
from nadir
𝑟𝑠𝑢𝑏 ∗ ℎ𝑠𝑢𝑏
3.73" ∗ (+119.09′)
=
=
= +0.03701"
𝐻𝑛𝑎𝑑𝑖𝑟
12000′
Terrain displacement is
extremely slight on 1:24100
photos of ISU Campus
Correct map
position 0.037”
closer to nadir
Topographic Displacement:
All about Similar Triangles
𝒅=
𝒓𝒉
𝑯
We get: 𝒉 =
𝒅𝑯
𝒓
From:
Measurements
based on…
Photo # 24
𝒓
𝒓′
𝒅
H = 3000’
NADIR
𝑟 ′ = 22.45 * 0.2 = 4.49”
𝑟 = 22.95 * 0.2 = 4.59”
Carver Hall
𝑑 = 𝑟 − 𝑟′
𝑑 = 4.59" − 4.49 = 0.1"
NADIR
ℎ=
Ground
Note: inches cancel
out leaving h in feet.
0.1" ∗ 3000′
300′
=
= 65.36′ ≈ 65′
4.59"
4.59
Tilt Displacement: Radial from Isocenter
• Is this a truly “vertical” photograph?
• How can one tell?
• Where are the PP & Nadir points?
Displacement due to Tilt: 𝒅𝒕
If you take a photo using a 6” CFL with 3o of Tilt (𝒕), how much displacement is there in an
image if Y is 4”?
Principal Meridian
“Nose-up” side of tilt (horizon side)
in radians
(4")2
𝑑𝑡 =
6"
− 4"
0.0532
dt
“Isoline” or
tilt axis
PP
I
Y
𝑑𝑡 = 0.14" ⇒ Correct map position
actually 0.14” farther
from nadir on photo
N
displacement
Tilt Displacement
On a tilted photograph, the nadir point was determined by intersecting lines passing
through perfectly tall and clearly visible features on the photograph.
The distance between the nadir & the principal point was measured to be 0.5 inches.
What was the angle of tilt of the camera at the time of exposure if a 6 inch CFL lens was
used?
|𝑃𝑃 − 𝑛𝑎𝑑𝑖𝑟|
𝑇𝑖𝑙𝑡 𝐴𝑛𝑔𝑙𝑒 𝜃 = 𝑡𝑎𝑛−1
𝐶𝐹𝐿
𝑇𝑖𝑙𝑡 𝐴𝑛𝑔𝑙𝑒 𝜃 = 𝑡𝑎𝑛−1
0.5 𝑖𝑛.
6 𝑖𝑛.
= 4.764𝑜
Nothing in radians!
Another Topographic Displacement Problem
What can be done to make displacement as small as possible?
• The smaller the flying height or CFL, the greater the displacement.
• The taller the object (or deeper the hole) the more serious the displacement.
• Displacement is more serious the further we move from the nadir.
Suppose we fly at 13,750ft. and photograph a level terrain with a single 200ft. hill
(above datum elev.)…
How serious is photo displacement if 𝒓 = 𝟐. 𝟐𝟓“ and scale is 1:20,000?
𝒅=
𝒓𝒉
𝑯
=
2.25" ∗ 200𝑓𝑡.
13,750𝑓𝑡.
= 0.033“ …on the photo. What’s the ground distance?
1′
𝐺𝐷 = 0.033" ∗ 20,000 ∗
= 55′
12"
Another Height Problem
Rearrange displacement formula to get height formula as before…
𝑑=
𝑟ℎ
𝐻
⇒
ℎ=
𝑑𝐻
𝑟
NOTE: 𝒉 formula only works when 𝒅 (i.e., 𝒓 & 𝒓′) is actually visible on the photo
Suppose we are using photos at a scale of 1:10,000.
The flying height for the photos was 10,000 feet. The base of a building can be
seen 3.16 inches (𝑟′)from the nadir. The top of the same building is 3.18 inches (𝑟)
from the nadir.
What is the height of the building?
𝒉=
𝒅𝑯
3.18"−3.16" ∗ 10000𝑓𝑡.
=
= 63𝑓𝑡.
𝒓
3.8"
Area Determination Error Due to TILT
A
B
Above Datum…
Area is over estimated
Below Datum…
Area under estimated
Solid lines around A & B = true map positions of 2000-feet square areas at MSR = 12,000 on a single
photo…
Dashed lines show photo position & shape of same areas on PSR = 12,000 photo
•
Area A is 600’ higher than Nadir elevation (1000’)
• Here, PSR can be re-calculated & correct areas can be obtained
•
Left extreme of area B is 600’ above nadir elev. & right extreme is 600’ below Nadir elev.
• No mathematical correction is possible for average scale difference.
• In B, can only estimate area accurately near the axis of tilt
• If one of the photos in a stereo pair captures “your area” near Nadir, use that photo.
Effects of Topo Displacement on Area Determination
A more subtle example….
Week 4
Wednesday
Height Determination: The Shadow Method
2 methods for measuring heights on a single aerial photo:
1) Topographic displacement method
•
•
•
•
Object being measured must be vertical (top  bottom)
Object must be far enough from nadir to see displacement
Photo scale must be large enough to reliably measure displacement
Must be able to see both top & bottom of the object
2) Sun-Angle Shadow method
•
•
•
•
•
Must be able to measure full length of object’s shadow
Object’s top must be sharply pointed/distinct and transfers to shadow
Must know sun elevation angle above horizon
Must know the precise time the photo was taken
The rest is simple trigonometry to get object height (ℎ)
Actually,
somewhat
rare!
Height Determination: The Shadow Method
Not a trivial task to determine sun elevation angle!
1) Need the specific Lat/Lon of the object you wish to measure…
•
•
Can be interpolated from 1:24000, 7.5 minute Topoquad maps
Can also use Google Earth or other GIS sources
2) Need the exact time of photo exposure
•
•
Photos we use in class have this info clipped off
The original still has it…
3) Need to plug time and location (Lat/Lon) into a Sun Elevation Calculator
or use solar altitude tables is info above is known.
•
http://www.esrl.noaa.gov/gmd/grad/solcalc/azel.html
Height Determination: The Shadow Method
Perfect!
1) Vertical
2) Pointy top
Shadow correct
Shadow long
Shadow long
Instances that lead to error
in height estimation using
tree shadows
Shadow short
Shadow short
Shadow long
Shadow top
not resolved
on photo
Brush or snow
Shadow short
Shadow short
Height Determination: The Shadow Method
So, if all object conditions are met and we have all the photo ephemeris info…
Then height by shadow measurement is a simple process:
Sun Elevation Angle (𝜃)
Object Height
Shadow Length
𝐹𝑖𝑟𝑠𝑡, 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝜽 𝑓𝑟𝑜𝑚 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑡𝑜 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
𝑂𝑏𝑗𝑒𝑐𝑡 𝐻𝑒𝑖𝑔ℎ𝑡
𝑇𝑎𝑛 𝜃 =
𝑆ℎ𝑎𝑑𝑜𝑤 𝐿𝑒𝑛𝑔𝑡ℎ
⇒
𝑑𝑒𝑔𝑟𝑒𝑒𝑠 ∗ 𝜋
180
𝐻𝑒𝑖𝑔ℎ𝑡 = 𝑇𝑎𝑛 𝜃 ∗ (𝑆ℎ𝑎𝑑𝑜𝑤 𝐿𝑒𝑛𝑔𝑡ℎ)
Shadow Method: Determining Object Height
1) Begin by measuring the photo length (PD) of the shadow and converting that
to a ground distance (GD) using the formula….
1
𝑃𝐷"
=
𝑃𝑆𝑅
𝐺𝐷"
⇒
𝑃𝑆𝑅 ∗ 𝑃𝐷" = GD“
1′
...GD" ∗
= 𝐺𝐷′
12"
2) Calculate the tangent of sun elevation angle & multiply by shadow length
Example:
1) You determine the shadow length of a radio tower is 200 ft.
2) You find that the sun elevation angle was 43o 
3) you find… 𝑇𝑎𝑛 0.750492 = 0.932515
4) With this, Height = 0.932515 ∗ 200 ft. = 𝟏𝟖𝟔. 𝟓 𝐟𝐭.
43 ∗ 𝜋
180
= 0.750492 radians
Shadow Method: Determining Sun Elevation
A more difficult problem is determining the sun’s elevation angle when an
object’s height and its shadow length are known.
For this, you can find sun's elevation angle by solving the height formula for
tangent of sun's elevation angle.
𝐻𝑒𝑖𝑔ℎ𝑡 = 𝑻𝒂𝒏 𝜽 ∗ (𝑆ℎ𝑎𝑑𝑜𝑤 𝐿𝑒𝑛𝑔𝑡ℎ)
For Example:
1) Suppose you know an object is 50 ft. tall and the shadow length = 38 ft.
2) 𝑇𝑎𝑛 𝑆𝑢𝑛′ 𝑠 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑔𝑙𝑒 =
3) This
=
50 𝑓𝑡.
38 𝑓𝑡.
𝐻𝑒𝑖𝑔ℎ𝑡
𝑆ℎ𝑎𝑑𝑜𝑤 𝐿𝑒𝑛𝑔𝑡ℎ
= 1.3158
4) Arctan(1.3158) = 0.92093 radians

0.92093 ∗180
𝜋
= 52.8𝑜
Determining Photo-Specific Solar Position
Most often, you won't know the height of objects casting shadows, so…
Options:
1) Can use a solar altitude table which is based on latitude (LAT) of the
photography & time the photo was taken.
•
Will give an approximate value for sun's altitude depending on how
closely LAT & time match those available in the table
•
It is possible to interpolate the table for time, date, & LAT to get an exact
answer for sun's altitude
2) Or, can use solar ephemeris to determine Sun’s altitude. To use ephemeris
must know the following:
•
•
•
•
Angle X ….Sun’s declination (LAT) on the day of the photo (look up in
ephemeris
Angle Y ….LAT of photography (find this on a map)
Angle Z ….is the hour angle, or difference in longitude (LON) between
position of the sun and locality of the photography (can be calculated)
𝑆𝑖𝑛 𝐴𝑙𝑡. 𝐴𝑛𝑔𝑙𝑒 = 𝐶𝑜𝑠𝑋 𝐶𝑜𝑠𝑌 𝐶𝑜𝑠𝑍 + 𝒐𝒓 − 𝑆𝑖𝑛 𝑋 𝑆𝑖𝑛𝑌
+ from March 21 – Sept. 23 in Northern Hemisphere
– from Sept. 24 – March 20 in Southern Hemisphere
FYI only
Determining Photo-Specific Solar Position (Option 1)
Sun Position Calculator: http://aa.usno.navy.mil/data/docs/AltAz.php
output
Determining Photo-Specific Solar Position (Option 1)
Height Determination: The Shadow Method
Very hard to read clock!
Requires use of the dreaded
Magnifying Comparator
The Shadow Method: In Presence of Slopes
When shadows fall on either an uphill or downhill slope, the method
𝑯𝒆𝒊𝒈𝒉𝒕 = 𝑻𝒂𝒏 𝜽 ∗ (𝑺𝒉𝒂𝒅𝒐𝒘 𝑳𝒆𝒏𝒈𝒕𝒉)
…will give incorrect height estimates unless adjustments are made.
An adjustment formula can be used if degree of slope is known …(via clinometer etc.)
𝑯𝒆𝒊𝒈𝒉𝒕 = 𝑺𝒉𝒂𝒅𝒐𝒘𝑳𝒆𝒏 ∗ 𝑪𝒐𝒔 𝑺𝒍𝒐𝒑𝒆 ∗ [𝑻𝒂𝒏 𝑺𝒖𝒏𝑬𝒍𝒆𝒗 + 𝒐𝒓 − 𝑻𝒂𝒏 𝑺𝒍𝒐𝒑𝒆 ]
+ if shadow falls uphill
− if shadow falls downhill
Angles in Radians 
𝜃∗𝜋
180
Example:
1:15840 photo is taken at 11:30 CST on August 20st at 40o N LAT . A tree casts a shadow on a
uphill face of a 35o slope. On the photo the shadow measures 0.03”, what is the tree’s height?
ℎ = 0.03" ∗ 𝐶𝑜𝑠 0.610851 ∗ 𝑇𝑎𝑛 1.064651 + 𝑇𝑎𝑛 0.610851
ℎ = 0.03" ∗ 0.81916 ∗ (1.804048 + 0.700186)
ℎ = 0.03" ∗ 2.051369
ℎ = 0.061541" ∗ (15840/12) = 𝟖𝟏. 𝟐′
≈ 𝟖𝟐′
Get sun elev. from SOLAR TABLE
Hotspot
If the sun's altitude is greater than 90 minus 1/2 the lens angle of the camera being used
to take the photos, rays from the sun can be reflected directly onto the film causing a sun
spot to appear on the photo
> 60o
60o to 52.5o
52.5o to 40o
< 40o
Hotspot
>
Sun's Image or Hot Spot
will cause loss of detail in a
small area. It appears as a
white circular spot on the
photo.
=
<
No Hotspot
𝜃 = 90 −
𝜃
1
∗ Lens Angle
2
Hotspot & No Shadow Spot
No Shadow Spot: same distance from the Principal
point as the Sun's Image along a line drawn from the
sun's image through the PP. No Shadow point will be
on the opposite side of PP from the sun's image
PP
No
Shadow
spot
Hotspot
No Shadow Spot
1)
No Shadow Spot: opposite side of PP from the sun's image (hotspot)
on a line running through the PP.
2)
Large area where no shadows will appear on the photo
•
Caused by the fact that the displaced images of objects fall
directly on top of the shadow being cast by the object.
•
Result is loss of detail over a large area & reduction in ability to
interpret aerial photos.
•
Ideally, would like to obtain photos in such a way that both sun
spot & no shadow spot are avoided.
VS.