basics of imaging systems lecture 3 prepared by rick lathrop 9/99 revised 9/06

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Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

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Page 1: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Basics of Imaging systemsLecture 3

prepared by Rick Lathrop 9/99

revised 9/06

Page 2: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Learning objectives• RSS concepts:

– Basic components/mechanics of image framing vs. scanning systems

– Concept of focal length– Variables affecting image exposure– Image scale and ground coverage and their relationship with

flying height and focal length

• Math concepts:– Scale equation(s) as fundamental to all of geomatics– Basic application of trigonometry

Page 3: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Framing systems Instantaneously acquire an image• Film Camera - uses a lens to form an image at

the focal plane. A shutter opens at selected intervals to allow light to enter, where the image is recorded on photographic film or an array of detectors

• Digital Camera - type of camera that records an image on an array of photosensitive electronically charged detectors that is recorded on magnetic disk

Page 4: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Components of a framing camera

• Lens - function is to gather light directed from the ground scene and bring it into focus at the focal plane

• Focal length - the linear distance from the center of the lens to the focal plane

• Shutter speed - various times of exposure

• Diaphragm - controls the amount of light transmitted to the film when the shutter is open

Page 5: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Components of a framing camera

Graphic from http://library.thinkquest.org/16541/eng/explore/media/photos/camera_diagram.jpg

Page 6: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Single-lens reflex camera

Text and graphics from http://en.wikipedia.org/wiki/Image:Slr-cross-section.png

This cross-section (side-view) of the optical components of an SLR shows how the light passes through the lens (1), is reflected by the mirror (2) and is projected on the matte focusing screen (5). Via a condensing lens (6) and internal reflections in the pentaprism (7) the image appears in the eye piece (8). When an image is taken, the mirror moves in the direction of the arrow, the focal plane shutter (3) opens, and the image is projected in the film (4) in exactly the same manner as on the

focusing screen

Page 7: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Focal length

o i

Graphic from http://en.wikipedia.org/wiki/Lens_%28optics%29

Page 8: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

1/f = 1/o + 1/i

• Where f = focal length o = object distance = object to lens i = image distance = lens to focal plane

• f is constant, as object distance changes the image distance must change. In aerial photos, o is large, 1/o goes to zero, so i must equal f

Page 9: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Mapping or metric camera

• Single lens frame camera• High geometric quality• Film format is 230 mm (~9 in)

on a side• Focal length of 152 mm

common• Fiducial marks for later

registration and defining principal point of the photo

Keystone’s Wild RC-10 mapping camera

B&W NAPP photo

Page 10: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Large Format Camera (LFC)•Large Format Camera (LFC) flown in Space Shuttle has f = 305 mm and film size of 230x460 mm which resulted in a typical ground dimensions of 225 x 450 km (140 x 280 mi)

Photos: NASA

Page 11: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Digital Framing/Scanning Systems

• Charge coupled device (CCD): electronic sensor sensitive to a particular wavelength of light, that are generally physically separate on the focal plane

• RGB color image generally has separate RGB CCDs

• There can be difficulty in spatial co-registering of the different wavebands for the same pixel

Page 12: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Digital Mapping Camera: Zeiss/Intergraph Imaging

•2d CCD matrix (array) to ensure a rigid image geometry similar to a traditional precision film platen

•Panchromatic 7000 x 4000 pixels •Color 3000 x 2000 pixels

•Separate lens for each band•Multiple smaller camera heads to create image rather than a single, large diameter •12 bit radiometric resolution

http://imgs.intergraph.com/dmc/

Page 13: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Digital Line Sensing Systems:Leica Airborne Digital Sensor (ADS40)

http://www.gis.leica-geosystems.com/products/ads40/

•Pushbroom linear array system rather than a 2D framing system•3 line scanners : forwards, downwards and backwards to provide for stereoscopic coverage•Three CCD sensors: B&W color (RGB) & NIR

12,000 pixels across •RGB co-registration through special trichroid filter that splits beam from single lens, rather than 3 different lens•Field of View of 64o

•Produces up to 100GB of data per hour of flight

Page 14: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Compact Airborne Spectrographic Imager (CASI)

• Hyperspectral: 288 channels between 0.4-0.9 m; each channel 0.018m wide

• Spatial resolution depends on flying height of aircraft

For more info: www.itres.com

CASI 550

Page 15: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Most aerial photo mapping missions require overlapped coverage of successive

aerial photos along a flight line

Page 16: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Pushbroom Scanning vs. 2D Framing

Graphics from http://www.gis.leica-geosystems.com/products/documents/ADS40_product_description.pdf

Page 17: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Film Exposure

Graphics from http://www.photoretouchingsecrets.com/imagefiles/

Overexposed Underexposed

Page 18: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Image Exposure

• Exposure, E = s * d2 * t / 4 f2

• whereE = image exposure, J mm-2

s = scene brightness, J mm-

2 sec-1 d = diameter of lens opening, mm t = exposure time, sec

f = focal length of lens, mm

Page 19: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Photo exposure example

Case 1 Case 2

f = 40 mm if d = 10 mm

d = 5 mm t = ?

t = 1/125 s

E1 = s1 (d1)2 t1 = s2 (d2)2 t2 = E2

4(f1)2 4(f2)2

Page 20: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Image exposure exampleCase 1 Case 2

f = 40 mm if d = 10 mm

d = 5 mm t = ?

t = 1/125 s

E1 = s1* (5)2* 1/125sec = s2* (10)2* t2 = E2

4(40)2 4(40)2

t2 = (5)2* 1/125sec = 25/125 sec = 1 / 500 sec

(10)2 100

Page 21: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

F/STOP

• F/STOP = relative aperture or lens opening

• F/STOP = f/d = lens focal length/ lens opening diameter

• F/STOP increases, d decreases, E decreases

• must change F/STOP and exp. time, t, together. As F/STOP increases, t increases

• E = s * t / 4 F2

Page 22: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

F/STOP example

Case 1 Case 2

F/8 if F/STOP = F/4

t = 1/125 s t = ?

E1 = s1 t1 = s2 t2 = E2

4(F1)2 4(F2)2

Page 23: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

F/STOP example

Case 1 Case 2

F/8 if F/STOP = F/4

t = 1/125 s t = ?

E1 = 1/125s = t2 = E2

4(8)2 4(4)2

t2 = 1/125s * (4)2 = 16/125 sec = 1/500 sec

(8)2 64

Page 24: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

F/STOPEach F/STOP changes the amount of light E by a factor of 2

F/STOP Shutter Speed

22 all equal 1/4 2x exp

16 2x exp 1/30 at F/8 1/8

11 1/60 at F/5.6 1/15

8 1/125 at F/4 1/30

5.6 another example 1/60

4 1/125 at F/8 1/125

2.8 1/250 at F/5.6 1/250

2 1/500 at F/4 1/500

1.4 1/1000

Page 25: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

F/STOP Interrelationships

• Double the focal length, quadruple the time

• 1/4 the time, double the diameter or 1/2 F/STOP

• double the f, double the diameter

Page 26: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Lens speed

• The larger the lens diameter at full aperture, the more light the lens will admit in a given time interval

• lens speed = F/STOP at full aperture• the smaller the F/STOP, the faster the lens

F/2 has double the aperture diameter as F/4• fast speed lenses are needed for low light

conditions

Page 27: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Example: F/Stop effect on depth of field

F/22

F/8

F/4

http://en.wikipedia.org/wiki/Depth_of_field

F/STOP increases, Depth of Field increases

The range of distance over which objects are in focus

Page 28: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Remote Sensing Platforms• Geostationary

• Polar orbit

• manned space

• High altitude aircraft (U-2)

• Jets

• low alt. aircraft

• Platforms

• In-situ/ground

36,000km

900 km

200-300 km

90,000 ft

10-30,000 ft

500-10,000 ft

10-100 ft

0-5 ft

Page 29: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Image Scale

• Scale = f /H’ = d/D • where

f = focal lengthH’ = height above terraind = image distanceD = ground distanceh = terrain elevationH = flying height (h + H’)

H’

f

D

d

h

H

Page 30: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Example: Image Scale vs. flying height example

• Scale = 1 /RFd = f /H’ = d/D

• where f = 152 mm

D = 5000m

d = 230mm

H’ = ?

H’ = f x D = 152mm x 5000m = 3304 m = 3300m

d 230 mm

If I want a ground coverage of 5km, what flying height should I use?

Page 31: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Example: Image Scale vs. flying height example

• Scale = 1 /RFd = 1/50,000 = f /H’ • where

f = 152 mm

H’ = ?

H’ = f x RFd = 152mm x 50,000 = 7600m 1 1

If I want a scale of 1/50,000, what flying height should I use?

Page 32: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Effect of flying height on ground coverage

x

Adapted from Lillesand & Kiefer, 2nd edition

H’1

H’2

H’1 > H’2

D1 > D2

D2

D1

Page 33: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Effect of focal length on ground coverage

x

Adapted from Lillesand & Kiefer, 2nd edition

H’1

f1 > f2

D1 < D2

f1

f2

D1

D2

Page 34: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Ground Coverage

• Ground coverage, D, of photo frame varies with f and H’

• as f decreases, ground coverage increasese.g. f1 = 1/2 f2 D1 = 2D2 A1 = 4A2

• as H’ increases, ground coverage increases e.g. H’2 = 2H’1 D2 = 2D1 A2 = 4A1

Page 35: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Ground Coverage example

Case 1 Case 2 film size = 9.0” = 230mm film size = 9.0” f1 = 210 mm f2 = 152 mm H’ = 12,200 m H’ = 12,200 m Scale = ? Scale = ? D = ? D = ?

Page 36: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Ground Coverage example

Case 1 film size = 9.0” = 230mm

f1 = 210 mm H’ = 12,200 m Scale = f / H’ = 210mm / 12,200m = .210m / 12,200m = 1 / 58,000 Scale = d / D D = d * Scale RFd D = 230mm x 58,000 = 13340000 mm = 13.34 km

Page 37: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

National High Altitude program (NHAP)

• Flying Height, H’ = 12,200 m• color IR camera

f = 210 mm scale 1:58,000 area per frame 13.3 x 13.3 km

• panchromatic camera f = 152 mm

scale 1:80,000 area per frame 18.4 x 18.4 km

Page 38: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Ground Coverage for Scanning Systems

• W = 2 H’ tan where W = swath width

H’ = flying height above terrain

= one half total field of view of scanner

H’

W

Hint: remember your trigonometry

Tangent of a right angleopposite

adjacent

Opposite = tan * adjacent

opp

adj

Page 39: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Ground Coverage for Scanning Systems

• W = 2 H’ tan • Example: Leica ADS40 = 64o

if H’ = 2880 m

W = 2 * 2880m * tan32o = 3600m

H’

W

Page 40: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Extra Puzzler

• The Quickbird satellite is flown at an altitude of 450 km, with a total angular field of view of 2.12o. What is the swath width?

Page 41: Basics of Imaging systems Lecture 3 prepared by Rick Lathrop 9/99 revised 9/06

Extra Puzzler

• The Quickbird satellite is flown at an altitude of 450 km, with a total angular field of view of 2.12o. What is the swath width?

• W = 2 H’ tan • W = 2 * 450km * tan (1.06o)

= 900km * 0.0185

• W = 16.65km