Distortion and Displacement
Distortion β ββ¦any shift in the position of an image on a photograph that alters the perspective characteristics of the image.β
Displacement β ββ¦any shift in the position of an image on a photograph that does not alter the perspective characteristics of the photographβ
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
Usually negligible
Lens distortion radiates from the PP
β’ Causes an image to appear closer to or farther from the PP β’ most serious near the edges of the photo
Distortion and Displacement
I. Principal Points (PP) & Conjugate Principal Points (CPP) β Not the only points of interest on an aerial photo
A) Nadir & Isocenter β also important, especially on oblique photos.
B) PP β 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 slightβ¦.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 distortionβ¦
β’ Can we accurately measure scale, distance, or area using this photo? β’ Is average PRS helpful for area calculation?
Displacement
On a vertical photograph over flat terrainβ¦
β’ The farther any vertical object is from nadirβ¦the greater the objectβs top is displaced away from nadir
β’ The objectβs bottom position remains stationary (i.e., no displacement)
However, Terrain Variation⦠⒠displaces both the top & bottom of
vertical objects.
β’ How much & direction (toward/away from Nadir) depends on magnitude of elev. β (+/-) w/respect to Nadir elev.
Distortion and Displacement
D) Isocenter β 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 Iowaβ¦why?)
β’ Images on βup sideβ of a tilted photo are displaced toward photo center
β’ Images on βdown sideβ 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 = Flying height above ground h = Height (+/β) of an object on the ground (tree, building, etc.) d = Displacement distance on photo (inches or mm) r = Radial distance on photo (inches or mm) from nadir to the displaced point A = Altitude of aircraft above MSL E = Elevation of ground level or datum
Displacement: All about Similar Triangles
Displacement ( d )
d = r β rβ Consider following relationships: f r as H-h R f rβ as H R
MSL
Since d = r β rβ We have to solve for r and rβ separatelyβ¦
In this Exampleβ¦
r = photo dist.: nadir to building top
rβ = photo dist.: nadir to building base
d = displacement dist.: top β base r β rβ
Topographic Displacement: All about Similar Triangles
π
π=
π» β β
π β΄ π =
π(π )
π» β β
π
πβ²=
π»
π β΄ πβ² =
π(π )
π»
π = π β πβ²
π =ππ
π»βββ
ππ
π»
π =π»(ππ )
π»(π»ββ)β
π»ββ ππ
π» π»ββ
π =π»(ππ )
π»(π»ββ)β
π»ππ ββππ
π» ββπ»
π =π»ππ β π»ππ + βππ
π»(π»ββ)
π =βππ
π»(π»ββ)
π =πβ
π»
r
1
2
π = π β πβ²
Topographic Displacement at ISU
NADIR
E = 915.98β
E = 1035.07β hsub = +119.09β
E = 876.05β hriv = -39.93β
rriv nadir to river 13.15 * 0.2 = 2.63β
rsub nadir to suburb 18.65 * 0.2 = 3.73β
ππππ£ = ππππ£ β βπππ£
π»πππππ=
2.63" β (β39.93β²)
12000β²= β0.00875"
ππ π’π = ππ π’π β βπ π’π
π»πππππ=
3.73" β (+119.09β²)
12000β²= +0.03701"
π«πππππππππππ: π =ππ
π―
Suburb
River
NADIR Measurements based onβ¦ Photo # 30 Scale 1:24100 H = 12000β
Correct map position 0.037β closer to nadir
Actual Actual
Displacement
Terrain displacement is extremely slight on 1:24100 photos of ISU Campus
Correct map position 0.00875β farther away from nadir
Topographic Displacement: All about Similar Triangles
πβ² = 22.45 * 0.2 = 4.49β
π = 22.95 * 0.2 = 4.59β π = π β πβ²
π = 4.59" β 4.49 = 0.1"
β = 0.1" β 3000β²
4.59"=
300β²
4.59= 65.36β² β 65β²
From: π =ππ
π―
We get: π =π π―
π
NADIR
Measurements based onβ¦ Photo # 24 H = 3000β
Carver Hall
πβ² π
Note: inches cancel out leaving h in feet.
NA
DIR
Ground
π
β’ Is this a truly βverticalβ photograph? β’ How can one tell? β’ Where are the PP & Nadir points?
Tilt Displacement: Radial from Isocenter
Displacement due to Tilt: π π
βNose-upβ side of tilt (horizon side)
N
PP
I
Pri
nci
pa
l Mer
idia
n
βIsolineβ or tilt axis
dt Y
If you take a photo using a 6β CFL with 3o of Tilt (π), how much displacement is there in an image if Y is 4β?
ππ‘ = (4")2
6"0.0532
β 4"
ππ‘ = 0.14" β
in radians
Correct map position actually 0.14β farther from nadir on photo
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?
ππππ‘ π΄ππππ π = π‘ππβ1 0.5 ππ.
6 ππ. = 4.764π
Nothing in radians!
ππππ‘ π΄ππππ π = π‘ππβ1 |ππ β πππππ|
πΆπΉπΏ
Another Topographic Displacement Problem
What can be done to make displacement as small as possible?
β’ The smaller the flying height or CFL, the greater the displacement.
β’ The taller the object (or deeper the hole) the more serious the displacement.
β’ 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.)β¦ How serious is photo displacement if π = π. ππβ and scale is 1:20,000?
π =ππ
π―=
2.25" β 200ππ‘.
13,750ππ‘.= 0.033β β¦on the photo. Whatβs the ground distance?
πΊπ· = 0.033" β 20,000 β 1β²
12"= 55β²
Another Height Problem
Rearrange displacement formula to get height formula as beforeβ¦
π =πβ
π» β β =
ππ»
π
NOTE: π formula only works when π (i.e., π & πβ²) 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 (πβ²)from the nadir. The top of the same building is 3.18 inches (π) from the nadir. What is the height of the building?
π =π π―
π=
3.18"β3.16" β 10000ππ‘.
3.8"= 63ππ‘.
Solid lines around A & B = true map positions of 2000-feet square areas at MSR = 12,000 on a single photoβ¦
Dashed lines show photo position & shape of same areas on PSR = 12,000 photo β’ Area A is 600β higher than Nadir elevation (1000β)
β’ Here, PSR can be re-calculated & correct areas can be obtained
β’ Left extreme of area B is 600β above nadir elev. & right extreme is 600β below Nadir elev. β’ No mathematical correction is possible for average scale difference. β’ In B, can only estimate area accurately near the axis of tilt β’ If one of the photos in a stereo pair captures βyour areaβ near Nadir, use that photo.
Above Datum⦠Area is over estimated
Below Datum⦠Area under estimated
Area Determination Error Due to TILT
A B
Height Determination: The Shadow Method
2 methods for measuring heights on a single aerial photo: 1) Topographic displacement method
β’ Object being measured must be vertical (top bottom) β’ Object must be far enough from nadir to see displacement β’ Photo scale must be large enough to reliably measure displacement β’ Must be able to see both top & bottom of the object
2) Sun-Angle Shadow method β’ Must be able to measure full length of objectβs shadow β’ Objectβs top must be sharply pointed/distinct and transfers to shadow β’ Must know sun elevation angle above horizon β’ Must know the precise time the photo was taken β’ The rest is simple trigonometry to get object height (β)
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 measureβ¦
β’ Can be interpolated from 1:24000, 7.5 minute Topoquad maps β’ Can also use Google Earth or other GIS sources
2) Need the exact time of photo exposure β’ Photos we use in class have this info clipped off β’ The original still has itβ¦
3) Need to plug time and location (Lat/Lon) into a Sun Elevation Calculator
or use solar altitude tables is info above is known. β’ http://www.esrl.noaa.gov/gmd/grad/solcalc/azel.html
Height Determination: The Shadow Method
Instances that lead to error in height estimation using tree shadows
Perfect! 1) Vertical 2) Pointy top
Shadow correct Shadow long Shadow long
Shadow short Shadow short Shadow long
Shadow top not resolved on photo
Shadow short Shadow short
Brush or snow
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:
Shadow Length
Object Height Sun Elevation Angle (π)
πππ π = ππππππ‘ π»πππβπ‘
πβππππ€ πΏππππ‘β π»πππβπ‘ = πππ π β (πβππππ€ πΏππππ‘β)
πΉπππ π‘, ππππ£πππ‘ π½ ππππ πππππππ π‘π πππππππ β πππππππ β π
180
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 formulaβ¦.
1
πππ =
ππ·"
πΊπ·" β πππ β ππ·" = GDβ ...GD" β
1β²
12" = πΊπ·β²
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 43 β π
180= 0.750492 radians
3) you findβ¦ πππ 0.750492 = 0.932515
4) With this, Height = 0.932515 β 200 ft. = πππ. π ππ.
Shadow Method: Determining Sun Elevation
A more difficult problem is determining the sunβs elevation angle when an objectβs 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.
π»πππβπ‘ = π»ππ π½ β (πβππππ€ πΏππππ‘β)
1) Suppose you know an object is 50 ft. tall and the shadow length = 38 ft.
2) πππ ππ’πβ²π ππππ£ππ‘πππ πππππ = π»πππβπ‘
πβππππ€ πΏππππ‘β
3) This = 50 ππ‘.
38 ππ‘. = 1.3158
4) Arctan(1.3158) = 0.92093 radians 0.92093 β180
π= 52.8π
For Example:
Most often, you won't know the height of objects casting shadows, soβ¦
Options: 1) Can use a solar altitude table which is based on latitude (LAT) of the
photography & time the photo was taken.
β’ Will give an approximate value for sun's altitude depending on how closely LAT & time match those available in the table
β’ 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 Sunβs altitude. To use ephemeris
must know the following:
β’ Angle X β¦.Sunβs declination (LAT) on the day of the photo (look up in ephemeris
β’ Angle Y β¦.LAT of photography (find this on a map) β’ Angle Z β¦.is the hour angle, or difference in longitude (LON) between
position of the sun and locality of the photography (can be calculated)
β’ πππ π΄ππ‘. π΄ππππ = πΆππ π πΆππ π πΆππ π + ππ β πππ π ππππ
+ from March 21 β Sept. 23 in Northern Hemisphere
β from Sept. 24 β March 20 in Southern Hemisphere
Determining Photo-Specific Solar Position
FYI only
Determining Photo-Specific Solar Position (Option 1)
Sun Position Calculator: http://aa.usno.navy.mil/data/docs/AltAz.php
output
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
π―πππππ = π»ππ π½ β (πΊπππ ππ π³πππππ)
β¦will give incorrect height estimates unless adjustments are made.
An adjustment formula can be used if degree of slope is known β¦(via clinometer etc.)
π―πππππ = πΊπππ πππ³ππ β πͺππ πΊππππ β [π»ππ πΊπππ¬πππ + ππ β π»ππ πΊππππ ] + if shadow falls uphill
β if shadow falls downhill Angles in Radians π β 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 0.03β, what is the treeβs height?
β = 0.03" β πΆππ 0.610851 β πππ 1.064651 + πππ 0.610851
β = 0.03" β 0.81916 β (1.804048 + 0.700186)
β = 0.03" β 2.051369
β = 0.061541" β (15840/12) = ππ. πβ² β ππβ² 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
π = 90 β 1
2 β Lens Angle
=
π
> < Hotspot No 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.
Hotspot & No Shadow Spot
PP
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
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 β’ Caused by the fact that the displaced images of objects fall
directly on top of the shadow being cast by the object.
β’ Result is loss of detail over a large area & reduction in ability to interpret aerial photos.
β’ Ideally, would like to obtain photos in such a way that both sun spot & no shadow spot are avoided.
VS.