kuliah 6 – geometri foto udara

Ge

om

etr

i F

oto

Ud

ara

?

Optical distortion:

• Caused by inferior camera constant, lenses, atmospheric

interference etc.

• Of minor importance in modern aerial photography.

Tilt of the focal plane:

• Caused by aircraft (platform) orientation.

Relief displacement:

• Caused by the terrain undulations.

• The amount of displacement depends on the height of the object

and the radial distance of the object from the image nadir.

• The most important source of positional error.

Central projection (Photo) versus orthogonal projection (Map)

Perspective and Projection

• sinar-sinar proyeksi yang

dalam hal ini adalah

sejajar/paralel dan

• skala konstan untuk setiap

objek

• Sinar-sinar proyeksi melalui suatu titik O,

yaitu pusat perspektif

• posisi relatif dan geometri objek yang

digambarkan tergantung pada lokasi dari

mana foto itu diambil.

O

Foto Udara (+)

Foto Udara (-)

Orthogonal projection Central Perspective projection

• Varied Scale

• Relief Displacement

Different

1. Size,

2. Shape

3. Location of Static

Objects

Different

1. Size,

2. Shape

3. Location of Static

Objects

Orthorectifying this photo is necessary for practical use since the effects of relief displacement must be removed.

• On a map we see a top view of

objects in their true relative

horizontal positions. On a

photograph, areas of terrain at the

higher elevations lie closer to the c

amera and therefore appear larger

than the corresponding areas lying

at lower elevations.

• The image of the tops of objects

appearing in a photograph are

displaced from the images of their

bases. This distortion is known as

relief displacement and causes any

obje ct standing above the terrain

to lean away from the principal

point of a photo radially.

(A) Perspective projection (B) Orthogonal Projection

- Radial displacement- Scale differences

1. Optical distortion- Disebabkan karena masalah kamera

Central projection (Photo) versus orthogonal projection (Map)

Distorsi (distortions) :

suatu perubahan kedudukan suatu gambar pada suatu foto (shift)

yang mengubah ciri-ciri perspektif gambar.

diakibatkan perubahan lokasi foto yang mengubah sifat dasar dari

foto.

Pergeseran (displacement) :

Suatu perubahan kedudukan suatu gambar pada suatu foto yang

tidak mengubah ciri-ciri perspektif gambar.

disebabkan oleh perubahan dalam ketinggian dari mana foto itu

diambil.

Tipe Distorsi Tipe pergeseran

1. Pengerutan film dan gambar

cetakan (Film and Print

Shrinkage)

2. Pembiasan berkas cahaya di

dalam atmosfer (Atmospheric

refraction of light rays)

3. Gerakan Gambar (Image

motion)

4. Distorsi lensa (Lens

Distortions)

5. Malfungsi kamera: shutter

malfunction, failure of the film-

flattening mechanism in the

camera focal plane

1. Lengkungan bumi (Curvature of

the Earth)

2. Kemiringan sumbu kamera (tilt)

3. Bersifat topografis atau relief ,

termasuk tinggi obyek

(Topography and relief)

Efek dari penyusutan film, pembiasan atmosfer dan kelengkungan bumi biasanya diabaikan dalam banyak kasus -pengecualian adalah proyek pemetaan yang tepat.

Dipengaruhi :

• Kuliatas film dan kertas cetak

• Perubahan suhu (panas atau dingin)

Perubahan kecil kira-kira 0.025 mm

Efek dari penyusutan film, atmosfer refraksi biasanya diabaikan dalam banyak kasus

Koreksi :

Where:

x is the corrected photocoordinate along the x-axis for a point a,

y is the corrected photocoordinate along the y-axis for a point a,

xc is the calibrated fiducial distance along x-axis,

yc is the calibrated fiducial distance along y-axis,

xf is the measured fiducial distance along x-axis,

yf is the measured fiducial distance along y-axis,

xm is the measured photocoordinate for point a along the x-axis,

ym is the measured photocoordinate for point a along the y-axis, and

xc/xf and yc/yf are the correction factors along x-axis and y-axis respectively.

Example :

Suppose that the calibrated distances between the fiducial marks on the camera are 23.25 cm along x-axis and 23.30 cm along y-axis. The corresponding distances measured on a photographic print from the same camera are 23.33 cm and 23.36 cm. If the photocoordinates, x and y, of a point measured on the print are 8.15 cm and 11.04 cm, what are the corrected photocoordinates of the point.

Solution: foreknown:xc = 23.25 cmxf = 23.33 cmxm = 8.15 cmyc = 23.30 cmyf = 23.36 cmym = 11.04 cm

question:x = ……?y = ……?

Pembiasan terbesar

dekat “ground surface”

karena kepadatan

atmosfer

Where :

Z0 = Flying height above geoid (sea level), in km

ZP = Mean terrain height above geoid (sea level), in km

c = camera constant (mm)

r = radius for a point (x’,y’) in the image (mm)

Akibat pergerakan kamera (atau wahana) ketika exposure, yang

mengakibatkan noda (smearing) dan kekaburan (blurring) pada FU.

Untuk interpretasi dan pemetaan yang baik pada FU, pergerakan

gambar kira-kira 0.05 mm (0.002 in).

Meskipun Pergerakan gambar 0.353 mm (0.014 in) masih dapat

digunakan.

Where:

M = the image motion (movement) on the photograph (in millimeters in

equation (3.1.) and in inches in equation (3.2.)

0.2778 = a constant, with units: meter hours per kilometer second (in equation (3.1.)

17.6 = a constant, with units: inch hours per mile second (in equation (3.2.))

V = the ground speed of the plane in kilometers per hour in equation (3.1.) and

in miles per hour in equation (3.2.)

t = the shutter speed in seconds

f = the focal length of the camera lens (in mm in equation (3.1.) and in feet in

equation (3.2.))

HD = the flying height of the aircraft above the datum (in meters in equation (3.1.)

and in feet in equation (3.2.).

3.1.

3.2.

In both equations the term f/HD corresponds to the photo scale.

Therefore, the equations above may be rewritten as:

Where PSR is the photo scale reciprocal (1/photo scale or 1/(f/HD)).

Example :

Suppose that an airplane was flying 3000 meters (9840 feet) above the ground at 500 kilometers (about 310 miles) per hour. Suppose also that the camera was taking photographs with a 305-mm (12-in.) focal length camera lens and a shutter speed of 1/40th of a second (0.025 s). What would be the image motion?

foreknown:H = 3000 mv = 500 km/hrf = 305 mmt = 0.025 s

question:M = ……?

Solution:

( 0.025 s )

Berdasar rumus gerak gambar (persamaan 3.1. Atau 3.2.), ada banyak

cara untuk mengurangi gerakan gambar atau blur, yaitu dengan :

1. Menggunakan shutter speed yang lebih cepat (t)

2. Menggunakan pesawat terbang lebih lambat (v)

3. Terbang pada ketinggian yang lebih tinggi (hd) di atas tanah

4. Menggunakan panjang fokus lensa (f) yang lebih pendek (f)

Masalah gerak gambar kadang-kadang membuat sulit dihindari pada foto

skala besar, terutama ketika menggunakan pesawat cepat dan film warna

yang memiliki kecepatan film yang relatif lambat.

Example :

Suppose this time that an aircraft was flying at a speed of 450 km (about 280 miles) per hour and taking photographs with a shutter speed of 1/125th of a second using a focal length of 152.4 mm (6-in.). What should be the flying altitude of the aircraft above the ground to assure the acceptable image motion of 0.05 mm (0.002 in) on the photographs?

Solution:

Notice that the two results are slightly different due simply to data conversion between the English and the metric systems. In fact, 450 km = 279.6768 miles (not 280) and 0.05 mm = 0.0019685 in (not 0.002 in). If we use these two values (279.6768 milesand 0.0019685 in instead of 180 miles and 0.02 in ), we will find the exact number (about 10,000 ft) as in the metric equation. This is another indication that your measurements and your input data (aircraft speed, shutter speed, and focal length) must be as accurate as possible to obtain reliable and satisfactory photographs

Koreksi sistematis terhadap distorsi lensa menurut the Bureau of Standards USA :

< 0.01 mm (0.0004 inch) pada bagian tepi FU

Efek dari penyusutan film, pembiasan atmosfer dan

kelengkungan bumi biasanya diabaikan dalam banyak kasus -

pengecualian adalah proyek pemetaan yang presisi.

Distorsi (distortions) lensa ini biasanya efeknya kecil.

Pergeseran (displacement) biasanya masalah / efek

terbesar mempengaruhi analisis.

adalah pergeseran bayangan karena kelengkungan

bumi yang arahnya radial menuju ke titik nadir.

H’ . r3

Dr = --------------

2.R.f

Dr = Kelengkungan bumi

H’ = tinggi terbang

f = fokus kamera

R = jari-jari bumi

r = jarak radial antara bayangan dan titik nadir

Dimana :

The object space coordinate system (Ground Coordinate System) used in

photogrammetric formulas is a cartesian orthogonal right-hand system

with Z upwards.

However, control point coordinates often refer to some geodetic system,

where Z is the height relative to the sea level, geoid, ellipsoid or some

other curvedsurface.

There are different ways to deal with this problem;

1. Transform geodetic ground control point (GCP) coordinates to

(orthogonal) geocentric coordinates.

2. For limited areas, use tangent plane corrections for GCP Z-

coordinates and leave XY as they are.

3. Apply Earth curvature correction to image coordinates (or to the model

coordinates, but not as common).

a. An error in the position of a point on the photograph due to

indeliberate tilting of the aircraft

b. Due to instability of aircraft

c. May be due to tilting of the aircraft along the flight line

and/or perpendicular to the flight line

d. Increases radially from the isocenter

Gangguan kedudukan kamera karena kedudukan posisi pesawat.

Berubahnya ujud hipotetik yang berupa petak-petak bukursangkar seperti

pada gambar.

Airplane attitude is based on relative positions of the nose and wings on the natural horizon.

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?

Menentukan nadir yang menggunakan perpanjangan-perpanjangan

sisi-sisi gedung vertikal yang tinggi

Variasi skala

Rotasi terhadapsumbu Z

Rotasi terhadapsumbu X

Rotasi terhadap

sumbu Y Rotasi terhadapsumbu X& Y

Rotasi terhadap

sumbu X,Y& Z

Rotasi terhadap

sumbu X,Y,Zdan skala

kappa

phi

omegax

y

z

x

y

z

x

y

z

x

y

z

DISTORSI FOTO UDARAAkibat Pergerakan Pesawat

Koreksi/cara mengatasi :

Menggunakan sistem giroskop (gyroscopic system) pada sensor

untuk mengatasi roll.

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ASIMO untuk berjalan dengan dua kaki . Sistem kontrol ini yang memungkinkan robot

ASIMO untuk berjalan, berlari dan bahkan melompat sambil mempertahankan

stabilitasnya, sistem ini didasari kesadaran postural yang menjaga keseimbangan,persis

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Teknologi gyroscopic ini berperan penting dalam pengembangan sistem kontrol gerakan

untuk motor Honda MotoGP .

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Roll distortion

- about its flight axis

- roll compensation

Crab distortion

- caused by deflection of aircraft due to crosswind

- corrections: on the plane or by computer

Pitch distortion

- result in local scale change

- can be ignored in most analyses

Phototilt (t) • Amount of tilt of the aircraft (and

thus the camera lens) with respect

to the vertical axis

• Angle of tilt between the line

perpendicular to the horizontal

datum and the line perpendicular

to the lens

Where:

t = phototilt

Sa = scale of first point, projected to the principal line

Sb = scale of second point, projected to the principal line

y = distance between a and b along the principal line

Hmge = flying height with respect to the mean ground

Yang dimaksud dengan pergeseran relief adalah pergeseran posisibayangan suatu titik di atas foto yang disebabkan karena adanyaketinggian titik obyek di atas bidang datum.

Pergeseran relief pada gambar :• Pergeseran posisi p’p disebut pergeseran relief.• Arah pergeseran ini radial menjauhi pusat foto

karena titik p terletak di atas bidang datum.Sebaliknya untuk titik Q yang terletak dibawahbidang datum bayangannya adalah q sehinggapergeseran reliefnya q’q yang arahnya radialmenuju ke pusat foto.

Radial Displacement

Pergeseran relief pada foto vertikal

: Relief Displacement

Rel

ief

dis

pla

cem

ent

fro

m N

adir

(C

ente

r )

Efek height displacement pada gedung yang lebihtinggi

Relief Displacement increases with the radial distance.

Ayman F. Habib

Tinggi Objek (h) = …. ?Perhatikan ONA PQA

dimana :R = Jarak puncak ke dasar objekD = jarak puncak objek ke dasar objekH = Tinggi terbang

h = H*D/R

h/H = D/R

d

r

h

H

:

:

:

:

Pergeseran letak oleh relief pada foto /

Relief displacement (mm)

Jarak radial dari titik nadir ke obyek (mm)

Tinggi obyek di atas (+) atau di bawah (-)

bidang rujukan (m)

Tinggi terbang

Perhatikan AA’A” LOA”

Dimana :

D R -------- = --------

h H

Dengan menyatakan jarak D dan R pada skala FU akan diperoleh :

d r-------- = --------

h H

r . hd = ----------

H

Dari rumus ini harga pergeseraan relief akan bertambah besar bila :

a. jarak radial ( r ) dari titik nadir ( pusat foto vertikal ) bertambah besar.

b. ketinggian suatu titik terhadap datum (h) bertambah besar.

c. tinggi terbang makin rendah

• Relief Displacement is directly proportional to:

– Radial distance.

– Object height above the datum.

• Relief Displacement is inversely proportional to:

– Flying height above the datum.

Using this figure, determine

the height h of the building to

which are drawn white arrows

to distances d (photo

displacement from bottom to

top) and r (to building top). On

the actual photo (not your

screen) d = 0.5 inch and r =

3.0 inches. Scale of the photo

is 1:3600. Aircraft altitude is

1800 ft.

It is necessary first to convert

inches on the photo to feet on

the ground. Divide 3600 by

12, so that the scale can be

stated as 1 inch = 300 feet.

Then d becomes 150 ft and r

becomes 900 ft. Substituting

in the equation h = Hd/r =

(1800 x 150)/900 = 300 ft.

Contoh :

Jarak obyek yang tergambar pada foto ketitik nadir = 45 mm, tinggi terbang di atas bidangdatum = 3.000 m, tinggi obyek di atas bidang datum = 30 m. Berapakah pergeseran letak olehrelief dan ke arah mana ?

Diketahui :

r = 45 mm

h = 30 m

H = 3000 m

Ditanya :

∆r = ……?

Jawab :

Pada sebuah foto udara tegak dengan format baku terdapat gambar sebuahgedung bertingkat. Jarak antara titik tengah foto udara dengan dasar gedung76 mm, sedangkan jarak antara gambar puncak gedung ke titik tengah fotoudara tersebut adalah 81,46 mm. Tinggi terbang pesawat pemotret adalah1475 m, dan elevasi dataran tempat gedung berdiri adalah 427 meter.Tentukan berapa tinggi gedung tersebut.

Soal

Soal UjianAssume that the relief displacement for the summit of the tower is 5.3 mm (measured from the bottom, b, to the summit, s, of the tower on the photograph) and the radial distance measured from the photo center (assuming a true vertical photograph) to the base (b) of the tower is 59 mm. If the scale of the photograph is 1:10,000, as printed on the photograph, and the focal length used to take this photograph is 152.4 mm, how tall is the tower?

Solution:From equation 7.14, we notice that in order to solve for ho, we first need to determine the; flying height (HD) of the aircraft above the datum. The formula for the photo scale is f/HD.; Therefore, 1:10,000 = f/HD , thus: HD = 10,000 x f = 10,000 x 152.4 mm = 1524 m

Finally, substituting in equation 7.14, we obtain:

Variasi Pergeseran karena relief (Relief displacement), karena :

1. Ketinggian objek, semakin tinggi objek semakin besar relief displcement

2. Jarak objek dari titik Nadir, semakin jauh dari titik nadir semakin besar relief displacement

3. Ketinggian Terbang (semakin tinggi terbang semakin kecil relief displacement sehingga citra satelit di luar angkasa (H>>>705 km (Landsat))

Ad.1. Ketinggian Objek

2 (dua) menara yang tergambar pada FU benar-benar tegak diambil dari 2500 m dpal. Jarakpuncak masing-masing menara ke titik nadir FU sama yaitu 8.35 cm. Jika ketinggian menarapertama adalah 120 m dan ketinggian menara 2 adalah 85 m.Berapakah besarnya relief displacement masing-masing menara pada FU tersebut. Berikesimpulan dari hasil perhitungan.

Diketahui :H = 2500 mr1 = 8,35 cmr2 = 8,35 cmh1 = 120 mh2 = 85 m

Ditanya :d1 = ……?d2 = ……?

Contoh :

Kesimpulan :semakin tinggi objek semakin besar relief displacement

Jawab :r . h

d = ----------H

8,35 cm . 120 md1 = ------------------------

2500 m

d1 = 0,40 cm

8,35 cm . 85 md2 = ------------------------

2500 m

d2 = 0,29 cm

h1 > h2 → d1 > d2

Ad.2. Jarak objek dari titik Nadir

d

r

f

Negative Image

H-ho

HD

Contoh :

Menara pertama dan menara kedua mempunyai ketinggian yang sama 100 m di atas bidangdatum. Jarak puncak menara pertama ke titik nadir 6,55 cm, sedangkan jarak puncak menaradua ketitik nadir 9,21 cm. Ketinggian terbang adalah 2500 m. Hitunglah relief displacement masing-masing menara tersebut. Berikan kesimpulan yang Anda peroleh.

Diketahui :r1 = 6,55 cmr2 = 9,21 cmh1 = 100 mh2 = 100 mH = 2500 m

Ditanya :d1 = ……?d2 = ……?

Kesimpulan :Semakin jauh dari titik nadir semakin besar relief displacement yang terjadi

Jawab :r . h

d = ----------H

6,55 cm . 100 md1 = -------------------------

2500 m

d1 = 0,262 cm

9,21 cm . 100 md2 = --------------------------

2500 m

d2 = 0,368 cm

r1 < r2 → d1 < d2

Ad. 3. Ketinggian Terbang (semakin tinggi terbang semakin kecil relief displacement sehingga citra satelit di luar angkasa (H>>>705 km (Landsat))

Contoh :

Pada sebuah FU tergambar menara A yang mempunyai ketinggian 50 meter terukur jarak puncakmenara ke titik nadir 5 cm, diambil pada ketinggian terbang 500 m. Sedangkan pada citrapenginderaan jauh yang lain menara A terukur jarak puncaknya ke titik nadir sama yaitu 5 cm tetapi diambil oleh sebuah wahana dengan ketinggian 750 km. Hitunglah relief displacement masing-masing menara tersebut. Berikan kesimpulan yang Anda peroleh.

Diketahui :r1 = 5 cmr2 = 5 cmh1 = 50 mh2 = 50 mH1 = 500 mH2 = 750 km = 750000 m

Ditanya :d1 = ……?d2 = ……?

Kesimpulan :Semakin tinggi, tinggi terbang semakin kecil relief displacement yang terjadi

Jawab :r . h

d = ----------H

5 cm . 50 md1 = -------------------------

500 m

d1 = 0,5 cm

5 cm . 50 md2 = --------------------------

750000 m

d2 = 0,0003 cm = 0,003 mm

H1 < H2 → d1 > d2

Contoh :

(H>>>705 km (Landsat))

On an island, with a height h = 20 m above sea level, there is a

lighthouse on the highest point. An image is taken from an

altitude of 800 m above sea level. In the image we measure the

radius r’B = 54 mm to the base B’ of the lighthouse, and the length

of the radial displacement (along the vertical edge of the

lighthouse) Δr’= 2.4 mm. How high above the sea level is the top

of the lighthouse?

h = ....?

Occluded Area

The primary geometric distortion in vertical aerial photographs

is due to relief displacement. Objects directly below the centre

of the camera lens (i.e. at the nadir) will have only their tops

visible, while all other objects will appear to lean away from the

centre of the photo such that their tops and sides are visible. If

the objects are tall or are far away from the centre of the photo,

the distortion and positional error will be larger.